This deck assembles the full transcendental toolkit. It derives sine and cosine from the two special limits, rebuilds the other four trigonometric functions with the quotient rule, and covers the natural exponential that is its own derivative, general bases with their natural-log factor, the natural and general logarithm, and the inverse trigonometric derivatives - then combines all of it with the chain, product, and quotient rules. It targets the sign error on cosine and the co-functions, the degrees-versus-radians error, the missing natural-log factor on a general exponential, and the confusion between the logarithm and the exponential rules.
Subject: Calculus I · 149 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 09
Where these rules come from, why the minus signs land where they do, and how to chain them together.
Objectives
This deck finishes your differentiation toolkit. After it, you can differentiate essentially any function a Calculus I course will hand you.
Warm-up
Discussion prompt
Before we open Derivatives of Trig, Exponential, and Log Functions: without looking back, what was the main idea of The Chain Rule, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck is about differentiating a function built inside another function. It covers spotting the outer and inner pieces, both notations for the chain rule, and the generalized power rule, then works through trigonometric, exponential, and radical outer functions, double and triple compositions, chains inside the product and quotient rules, and chain values read from a table, closing with a preview of applied rates. It targets the classic errors of dropping the inner derivative, differentiating the inside in place, and misreading which piece is the inner function.
Section
Part 1
Concept
Every rule you have so far was built for powers of the variable, and for sums, products, and quotients of those.
\[ \frac{d}{dx}\left[x^{5}\right] = 5x^{4} \]
The sine function is not a power of anything. There is no exponent to bring down and nothing to subtract one from.
\[ \frac{d}{dx}\left[\sin x\right] = \; ? \]
So we go back to the one tool that always works, no matter how strange the function: the limit definition of the derivative.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Counterexample
Discussion prompt
Every rule you have so far was built for powers of the variable, and for sums, products, and quotients of those.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The sine function is not a power of anything. There is no exponent to bring down and nothing to subtract one from.
Picture it
Figure (svg): The sine curve over one full period with dashed tangent segments drawn at the start, the peak, the middle crossing, and the trough.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Before any algebra, guess the answer by walking along the sine curve and asking what the slope is doing.
Intuition
Before any algebra, guess the answer by walking along the sine curve and asking what the slope is doing.
Figure (svg): The sine curve over one full period with dashed tangent segments drawn at the start, the peak, the middle crossing, and the trough.
At the start the curve is climbing at its steepest. At the peak it levels off. Coming down through the middle it is falling at its steepest. At the trough it levels off again.
Now list those slopes in order: most positive, then zero, then most negative, then zero. That is exactly the shape of the cosine curve.
So the answer is almost certainly cosine. The rest of Part 1 is proving that the guess is exact, not just close.
Analogy
Discussion prompt
Explain Read the slopes off the sine graph first by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Before any algebra, guess the answer by walking along the sine curve and asking what the slope is doing.
Picture it
Animation
Shows: The sine curve, with its steepest and flattest points visible.
Where sine flattens, its slope is zero.
Takeaway: Where sine is steepest, cosine peaks; where sine levels off, cosine is zero. The rule is readable off the graph before it is proved.
Concept
Two limits do all the real work. Neither can be found by substitution, because both are the indeterminate form zero over zero.
\[ \lim_{h \to 0} \frac{\sin h}{h} = 1 \]
Watch it happen numerically. The angle is always in radians.
| h (radians) | sin h | sin h divided by h |
|---|---|---|
| 1 | 0.841471 | 0.841471 |
| 0.5 | 0.479426 | 0.958851 |
| 0.1 | 0.0998334 | 0.998334 |
| 0.01 | 0.00999983 | 0.999983 |
| 0.001 | 0.000999999833 | 0.99999983 |
The ratio is marching to one. Read the statement out loud as: for a small angle, the sine of the angle is nearly the angle itself.
Pattern
Step through it
Step through Special limit number one one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Figure (svg): A quarter of a unit circle with a radius drawn at a modest angle, the arc from the horizontal axis to that radius, and the vertical segment from the radius tip down to the axis.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
On a circle of radius one, the arc length cut off by an angle is literally the angle measured in radians. That is what radians mean.
Intuition
Figure (svg): A quarter of a unit circle with a radius drawn at a modest angle, the arc from the horizontal axis to that radius, and the vertical segment from the radius tip down to the axis.
On a circle of radius one, the arc length cut off by an angle is literally the angle measured in radians. That is what radians mean.
The dashed vertical segment is the sine of that angle. It is the straight-line shortcut across the same gap the arc curves around.
As the angle shrinks, the arc straightens out and the two lengths become indistinguishable. Their ratio squeezes to one.
\[ \cos h \;\le\; \frac{\sin h}{h} \;\le\; 1 \quad \text{for small } h \ne 0 \]
That sandwich is the honest proof: the outer two both go to one, so the middle is trapped and must go to one as well. This is the Squeeze Theorem doing its job.
Explain it
Discussion prompt
Explain Why the sine of a small angle is the angle to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
On a circle of radius one, the arc length cut off by an angle is literally the angle measured in radians. That is what radians mean.
Concept
The second limit is the one that quietly disappears from the final answer.
\[ \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \]
| h | cos h | (cos h - 1) divided by h |
|---|---|---|
| 0.1 | 0.9950042 | -0.049958 |
| 0.01 | 0.99995000 | -0.0050000 |
| 0.001 | 0.99999950 | -0.00050000 |
| -0.01 | 0.99995000 | +0.0050000 |
Cosine leaves its maximum value of one so slowly that the gap shrinks faster than the angle does. The ratio collapses to zero.
Said another way: the cosine curve has a horizontal tangent at the top. Special limit number two is that fact, written as a limit.
Pattern
Step through it
Step through Special limit number two one row at a time. What is driving the change, and what would the row after the last one be?
Ranking
Put in order
Put the moves of Worked example: the derivative of sine, from the definition into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The only way to separate the h from the x is to split the compound angle into pieces we can control.
Worked example
\[ \frac{d}{dx}\left[\sin x\right] = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \]
Expand the numerator with the angle-addition formula
Why: The only way to separate the h from the x is to split the compound angle into pieces we can control.
\[ \sin(x+h) = \sin x \cos h + \cos x \sin h \]
Substitute and group the terms that contain sine of x
Why: Both the first term and the subtracted term carry a factor of sine of x, so factor it out.
\[ \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} = \sin x\!\left(\frac{\cos h - 1}{h}\right) + \cos x\!\left(\frac{\sin h}{h}\right) \]
Pull the x-parts outside the limit
Why: The limit is taken as h approaches zero, so sine of x and cosine of x are constants here - they do not move.
\[ = \sin x \cdot \lim_{h \to 0}\frac{\cos h - 1}{h} \; + \; \cos x \cdot \lim_{h \to 0}\frac{\sin h}{h} \]
Substitute the two special limits
Why: The first special limit is zero, which kills the sine term. The second is one, which leaves cosine standing alone.
\[ = \sin x \cdot 0 + \cos x \cdot 1 = \cos x \]
Verify against the graph and a numerical slope
Why: At the peak the formula gives cosine of one quarter turn, which is zero, and the sine graph really is flat there. At the origin it predicts a slope of one, and the difference quotient with a step of 0.001 gives 0.99999983.
\[ \frac{d}{dx}\left[\sin x\right] = \cos x \]
Concept
\[ \frac{d}{dx}\left[\sin x\right] = \cos x \]
No minus sign, no extra factor, no rewriting. Sine differentiates to cosine, clean.
radian measure — The angle whose arc on a unit circle has length equal to the angle. Every derivative rule in this deck assumes the input is in radians.
Picture it
Animation
Shows: Arctangent flattens toward its asymptotes — a rendered Manim animation.
Rendered with Manim.
Takeaway: Its derivative peaks at the origin and dies at both ends — hence the flattening.
Picture it
Animation
Shows: Rule one: the derivative of sine — a rendered Manim animation.
Rendered with Manim.
Takeaway: Everything trigonometric is built from this one result.
Concept
Run the identical argument with the cosine angle-addition formula.
\[ \cos(x+h) = \cos x \cos h - \sin x \sin h \]
\[ \frac{\cos(x+h)-\cos x}{h} = \cos x\!\left(\frac{\cos h - 1}{h}\right) - \sin x\!\left(\frac{\sin h}{h}\right) \]
The same two special limits do the same job, but this time the minus sign in the identity survives.
\[ \frac{d}{dx}\left[\cos x\right] = -\sin x \]
The minus sign is not decoration. Cosine starts at its maximum and heads downward, so its derivative must start out negative.
Picture it
Animation
Shows: Rule two: the derivative of cosine — a rendered Manim animation.
Rendered with Manim.
Takeaway: Cosine falls where sine is positive — hence the minus sign in its rule.
Intuition
Keep differentiating sine and you come back to where you started after four steps.
\[ \sin x \;\longrightarrow\; \cos x \;\longrightarrow\; -\sin x \;\longrightarrow\; -\cos x \;\longrightarrow\; \sin x \]
This is a useful memory hook and a useful check. If you ever differentiate a sine or cosine four times and do not land back on the original, a sign got lost.
| derivative | of sine | of cosine |
|---|---|---|
| first | cos x | minus sin x |
| second | minus sin x | minus cos x |
| third | minus cos x | sin x |
| fourth | sin x | cos x |
Pattern
Step through it
Step through The four-step cycle one row at a time. What is driving the change, and what would the row after the last one be?
Picture it
Animation
Shows: Sine and cosine cycle every four derivatives — a rendered Manim animation.
Rendered with Manim.
Takeaway: Differentiate four times and you are back where you started.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Sine goes to cosine, so surely cosine goes to sine.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This claim predicts a slope of positive one there, because the sine of a quarter turn is one.
The identity for cosine carries a minus sign, and it survives the limit.
Why: This claim predicts a slope of positive one there, because the sine of a quarter turn is one.
Trap
Sine goes to cosine, so surely cosine goes to sine.
\[ \frac{d}{dx}\left[\cos x\right] \stackrel{?}{=} \sin x \]
Test it at a quarter turn
Why: This claim predicts a slope of positive one there, because the sine of a quarter turn is one.
\[ \text{claim: } \left.\frac{d}{dx}\cos x\right|_{x=\pi/2} = \sin\!\left(\tfrac{\pi}{2}\right) = +1 \]
Compare with the actual graph
Why: At a quarter turn the cosine curve is crossing zero on its way DOWN. A positive slope there is visibly impossible.
The identity for cosine carries a minus sign, and it survives the limit.
\[ \frac{d}{dx}\left[\cos x\right] = -\sin x \]
Test it at a quarter turn
Why: The rule predicts a slope of negative one, matching the downward crossing you can see on the graph.
\[ \left.\frac{d}{dx}\cos x\right|_{x=\pi/2} = -\sin\!\left(\tfrac{\pi}{2}\right) = -1 \]
Confirm numerically
Why: A symmetric difference quotient with a step of 0.001 gives negative 1.000000, not positive one.
\[ \frac{\cos(\tfrac{\pi}{2}+0.001) - \cos(\tfrac{\pi}{2}-0.001)}{0.002} = \frac{-0.0009999998 - 0.0009999998}{0.002} = -1.000000 \]
Notation
Annotate
From Trap: losing the minus sign on cosine — read this one piece at a time. What is each part doing?
On: \( \frac{d}{dx}\left[\cos x\right] \stackrel{?}{=} \sin x \)
Concept
Special limit number one is only true when the angle is measured in radians. Everything built on top of it inherits that condition.
In degrees the same ratio settles on a completely different number.
\[ \lim_{h \to 0} \frac{\sin(h^{\circ})}{h} = \frac{\pi}{180} \approx 0.0174533 \]
So the familiar rules are simply false for a degree input. Set the calculator to radians and leave it there for the rest of calculus.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Reading the angle as degrees and applying the rule anyway.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It predicts that sine climbs one unit for every one degree of angle - a slope of one.
Convert to radians first, then differentiate. The conversion is an inside function, so the chain rule leaves a factor behind.
Why: It predicts that sine climbs one unit for every one degree of angle - a slope of one.
Trap
Reading the angle as degrees and applying the rule anyway.
\[ \frac{d}{dx}\left[\sin(x^{\circ})\right] \stackrel{?}{=} \cos(x^{\circ}) \]
Test the claim at zero degrees
Why: It predicts that sine climbs one unit for every one degree of angle - a slope of one.
\[ \text{claim: slope at } 0^{\circ} = \cos(0^{\circ}) = 1 \]
Look at the actual numbers
Why: The sine of one degree is only 0.0174524, so over that one degree the function climbs by 0.017, not by 1. The claim is off by a factor of about 57.
Convert to radians first, then differentiate. The conversion is an inside function, so the chain rule leaves a factor behind.
\[ \sin(x^{\circ}) = \sin\!\left(\frac{\pi x}{180}\right) \]
Differentiate with the chain rule
Why: The outer sine becomes cosine, and the inner linear function contributes its constant slope.
\[ \frac{d}{dx}\left[\sin\!\left(\frac{\pi x}{180}\right)\right] = \frac{\pi}{180}\cos\!\left(\frac{\pi x}{180}\right) \]
Check the number at zero degrees
Why: The rule now predicts a slope of about 0.0174533 per degree, which matches the measured climb of 0.0174524 over the first degree.
Step zero
Discussion prompt
Worked example: a sine-and-cosine combination — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Split with the sum and constant-multiple rules
Answer:
Worked example
Find the derivative, then the slope of the tangent line at one third of a half turn.
\[ f(x) = 4\sin x + 3\cos x \]
Split with the sum and constant-multiple rules
Why: Differentiation goes through sums term by term, and a constant factor just rides along.
\[ f'(x) = 4\,\frac{d}{dx}\left[\sin x\right] + 3\,\frac{d}{dx}\left[\cos x\right] \]
Apply the two rules
Why: Sine becomes cosine with no sign change; cosine becomes sine WITH a sign change, and that minus turns the plus three into a minus three.
\[ f'(x) = 4\cos x + 3(-\sin x) = 4\cos x - 3\sin x \]
Evaluate at the stated angle
Why: Use the exact unit-circle values rather than a decimal, so the answer stays in test form.
\[ f'\!\left(\frac{\pi}{3}\right) = 4\cdot\frac{1}{2} - 3\cdot\frac{\sqrt{3}}{2} = 2 - \frac{3\sqrt{3}}{2} \approx -0.5981 \]
Verify with a numerical slope at the origin
Why: The formula predicts a slope of four at the origin. Measuring the real function there gives 3.985 over a step of 0.01, which is four to within the expected second-order error.
\[ \frac{f(0.01)-f(0)}{0.01} = \frac{3.0398493 - 3}{0.01} = 3.98493 \;\approx\; f'(0) = 4 \]
Picture it
Animation
Shows: Each line of the worked example "a sine-and-cosine combination", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts a slope of four at the origin. Measuring the real function there gives 3.985 over a step of 0.01, which is four to within the expected second-order error.
Ranking
Put in order
These are the steps of Playbook: any sine-and-cosine expression, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Five moves handle every sine-and-cosine derivative you will meet.
\[ \frac{d}{dx}\left[a\sin x + b\cos x\right] = a\cos x - b\sin x \]
Elimination
Eliminate the wrong options
What is the derivative of the function shown above?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The first term gives 2 cos x. The second term is minus five times cosine, and cosine differentiates to negative sine, so the two minus signs multiply to a plus: minus five times negative sine is plus five sine. The result is 2 cos x + 5 sin x.
Check
Differentiate this on paper before you choose. Watch the second term especially.
\[ f(x) = 2\sin x - 5\cos x \]
Check your understanding
What is the derivative of the function shown above?
Answer: A
Why: The first term gives 2 cos x. The second term is minus five times cosine, and cosine differentiates to negative sine, so the two minus signs multiply to a plus: minus five times negative sine is plus five sine. The result is 2 cos x + 5 sin x.
Section
Part 2
Concept
Tangent, cotangent, secant, and cosecant are all just sine and cosine in disguise.
\[ \tan x = \frac{\sin x}{\cos x}, \qquad \cot x = \frac{\cos x}{\sin x}, \qquad \sec x = \frac{1}{\cos x}, \qquad \csc x = \frac{1}{\sin x} \]
So the quotient rule plus the two rules you already have generates every remaining trig derivative. Memorize them for speed, but always be able to rebuild them.
Fill the middle
Fill in the blanks
From Worked example: rebuilding the tangent rule — finish the line. Write what belongs on the right of the equals sign before you look.
\frac\frac{d}{dx}\left[\frac{\sin x}{\cos x}\right]___\left[\tan x\right] = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The numerator order is bottom times the derivative of the top, MINUS top times the derivative of the bottom.
Worked example
\[ \frac{d}{dx}\left[\tan x\right] = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] \]
Set up the quotient rule with the top and bottom named
Why: The numerator order is bottom times the derivative of the top, MINUS top times the derivative of the bottom. Reversing it flips the sign of the whole answer.
\[ \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v u' - u v'}{v^{2}}, \qquad u = \sin x, \quad v = \cos x \]
Substitute the two derivatives
Why: The derivative of the bottom is negative sine, so the subtraction turns into an addition.
\[ = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^{2} x} = \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x} \]
Collapse the numerator with the Pythagorean identity
Why: Sine squared plus cosine squared is always one, which is why this rule comes out so clean.
\[ = \frac{1}{\cos^{2} x} = \sec^{2} x \]
Verify at the origin with a numerical slope
Why: The rule predicts a slope of one at the origin, since the secant of zero is one. Measuring the tangent function there gives 1.0000333 over a step of 0.01, which is one to within the expected error.
\[ \frac{d}{dx}\left[\tan x\right] = \sec^{2} x, \qquad \frac{\tan(0.01)-\tan(0)}{0.01} = \frac{0.010000333}{0.01} = 1.0000333 \]
Concept
\[ \frac{d}{dx}\left[\tan x\right] = \sec^{2} x \]
Run the identical quotient-rule computation on cotangent and the numerator comes out negative instead.
\[ \frac{d}{dx}\left[\cot x\right] = -\csc^{2} x \]
Both answers are squares, so both are never negative on their own. The only difference between the two rules is the minus sign out front.
That matches the graphs: every branch of tangent climbs, and every branch of cotangent falls.
Picture it
Animation
Shows: Tangent and secant — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both follow from the quotient rule on sine over cosine.
Estimation
Predict first
Secant is a reciprocal, so the chain rule with a negative-one exponent is faster than the quotient rule here.
Commit before you compute: what does Worked example: rebuilding the secant rule come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify at one sixth of a half turn
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The rule gives secant times tangent there, which is two times the square root of three, about 3.4641.
Worked example
Secant is a reciprocal, so the chain rule with a negative-one exponent is faster than the quotient rule here.
\[ \sec x = (\cos x)^{-1} \]
Apply the generalized power rule to the outer layer
Why: Bring the negative one down, subtract one from the exponent, and hold the inside unchanged for the moment.
\[ \frac{d}{dx}\left[(\cos x)^{-1}\right] = -1(\cos x)^{-2}\cdot \frac{d}{dx}\left[\cos x\right] \]
Multiply by the derivative of the inside
Why: The inside is cosine, whose derivative is negative sine. Two minus signs multiply to a plus.
\[ = -1(\cos x)^{-2}\cdot(-\sin x) = \frac{\sin x}{\cos^{2} x} \]
Split the fraction into recognizable pieces
Why: One of the cosines pairs with the sine to make tangent; the other becomes a secant.
\[ \frac{\sin x}{\cos^{2} x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x \]
Verify at one sixth of a half turn
Why: The rule gives secant times tangent there, which is two times the square root of three, about 3.4641. A symmetric difference quotient with a step of 0.001 gives 3.465, so the rule matches the real slope.
\[ \frac{d}{dx}\left[\sec x\right] = \sec x\tan x, \qquad \left.\sec x\tan x\right|_{x=\pi/3} = 2\sqrt{3} \approx 3.4641 \]
Picture it
Animation
Shows: Each line of the worked example "rebuilding the secant rule", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The rule gives secant times tangent there, which is two times the square root of three, about 3.4641. A symmetric difference quotient with a step of 0.001 gives 3.465, so the rule matches the real slope.
Concept
| function | derivative |
|---|---|
| sine | cosine |
| cosine | negative sine |
| tangent | secant squared |
| cotangent | negative cosecant squared |
| secant | secant times tangent |
| cosecant | negative cosecant times cotangent |
\[ \frac{d}{dx}\left[\csc x\right] = -\csc x \cot x, \qquad \frac{d}{dx}\left[\cot x\right] = -\csc^{2} x \]
Six rules, but really only two facts plus the quotient rule. If you blank on one under pressure, rebuild it.
Comparison
Comparison matrix
From The whole trig table at once: refill the derivative column from what you know. The rest of the table is as it appeared.
| function | derivative |
|---|---|
| sine | cosine |
| cosine | negative sine |
| tangent | secant squared |
| cotangent | negative cosecant squared |
| secant | secant times tangent |
| cosecant | negative cosecant times cotangent |
Intuition
Look down the derivative column. Every function whose name starts with co produces a negative derivative: cosine, cotangent, cosecant.
And the derivative of a co-function is built entirely out of co-functions. Sine gives cosine; cosine gives sine. Tangent gives secant squared; cotangent gives cosecant squared.
So you really only memorize three rules, then apply the mirror: add co to every name and attach a minus sign.
| memorized rule | mirror it to get |
|---|---|
| sine gives cosine | cosine gives negative sine |
| tangent gives secant squared | cotangent gives negative cosecant squared |
| secant gives secant tangent | cosecant gives negative cosecant cotangent |
Trade off
Comparison matrix
From The co-functions carry the minus sign: every row here is a choice with a cost. Fill the mirror it to get column, then say which row you would actually pick and what you give up for it.
| memorized rule | mirror it to get |
|---|---|
| sine gives cosine | cosine gives negative sine |
| tangent gives secant squared | cotangent gives negative cosecant squared |
| secant gives secant tangent | cosecant gives negative cosecant cotangent |
Fill the middle
Fill in the blanks
From Trap: a positive derivative for cotangent — finish the line. Write what belongs on the right of the equals sign before you look.
\frac} \csc^{2} x___\left[\cot x\right] \stackrel______}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. There the cosecant is the square root of two, so the claim predicts a slope of positive two.
Trap
Pattern-matching cotangent onto the tangent rule and keeping the sign.
\[ \frac{d}{dx}\left[\cot x\right] \stackrel{?}{=} \csc^{2} x \]
Test the claim at one quarter turn
Why: There the cosecant is the square root of two, so the claim predicts a slope of positive two.
\[ \text{claim: } \left.\frac{d}{dx}\cot x\right|_{x=\pi/4} = \csc^{2}\!\left(\tfrac{\pi}{4}\right) = 2 \]
Compare with the real values
Why: Cotangent at one quarter turn is 1, and a little later at 0.8 radians it is 0.9712 - it went DOWN. A positive slope of two is impossible.
Cotangent is a co-function, so its derivative carries the minus sign.
\[ \frac{d}{dx}\left[\cot x\right] = -\csc^{2} x \]
Test it at one quarter turn
Why: The rule predicts a slope of negative two, which agrees with the falling values.
\[ \left.-\csc^{2} x\right|_{x=\pi/4} = -2 \]
Confirm numerically
Why: Over a step of 0.01 the cotangent drops from 1 to 0.9801987, a measured slope of about negative 1.980, closing in on negative two.
\[ \frac{\cot(\tfrac{\pi}{4}+0.01)-\cot(\tfrac{\pi}{4})}{0.01} = \frac{0.9801987 - 1}{0.01} = -1.98013 \]
Translation
\( \left.-\csc^{2} x\right|_{x=\pi/4} = -2 \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Worked example
\[ y = x^{2}\sec x \]
Recognize a product, not a composition
Why: Two separate factors are multiplied, and neither one is inside the other, so the product rule is the tool.
Apply the product rule
Why: Derivative of the first times the second, plus the first times the derivative of the second.
\[ y' = 2x\sec x + x^{2}\cdot\frac{d}{dx}\left[\sec x\right] \]
Fill in the secant rule
Why: The derivative of secant is secant times tangent - a product, so it stays attached as one block.
\[ y' = 2x\sec x + x^{2}\sec x\tan x \]
Factor for a cleaner answer
Why: A common factor of x and secant sits in both terms; pulling it out is what a grader expects to see.
\[ y' = x\sec x\,(2 + x\tan x) \]
Verify numerically at one
Why: The formula gives about 6.5841 at that point. A symmetric difference quotient with a step of 0.001 gives about 6.5835, so the answer holds.
\[ y'(1) = \sec(1)\left(2 + \tan 1\right) = 1.85082\,(3.55741) \approx 6.5841 \]
Picture it
Animation
Shows: Each line of the worked example "a product with secant", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives about 6.5841 at that point. A symmetric difference quotient with a step of 0.001 gives about 6.5835, so the answer holds.
Check
Two factors are multiplied here. Decide which rule you need before you differentiate anything.
\[ f(x) = x\tan x \]
Check your understanding
What is the derivative of x times tangent x?
Answer: A
Why: This is a product. The derivative of x is 1, giving the first term tan x, and then x times the derivative of tangent gives x times secant squared. Together that is tan x plus x secant squared x.
Step zero
Discussion prompt
Worked example: a tangent line to the tangent curve — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Get the point on the curve first
Answer:
Worked example
Find the line tangent to this curve at one eighth of a full turn.
\[ y = \tan x \quad \text{at} \quad x = \frac{\pi}{4} \]
Get the point on the curve first
Why: A tangent line needs a point as well as a slope, and the point comes from the original function, not the derivative.
\[ y\!\left(\frac{\pi}{4}\right) = \tan\frac{\pi}{4} = 1 \]
Differentiate and evaluate for the slope
Why: The secant of one eighth of a turn is the square root of two, and squaring it gives exactly two.
\[ y' = \sec^{2} x, \qquad y'\!\left(\frac{\pi}{4}\right) = \left(\sqrt{2}\right)^{2} = 2 \]
Write the point-slope form
Why: Point-slope is the fastest correct form and needs no rearranging to be accepted.
\[ y - 1 = 2\left(x - \frac{\pi}{4}\right) \]
Verify the line hugs the curve
Why: Step 0.01 to the right of the point. The line predicts 1.02 and the true tangent value is 1.020204, so the line is within two ten-thousandths - exactly what a tangent should do.
\[ \tan\!\left(\frac{\pi}{4}+0.01\right) = 1.020204 \quad \text{vs.} \quad 1 + 2(0.01) = 1.02 \]
Picture it
Animation
Shows: Each line of the worked example "a tangent line to the tangent curve", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Step 0.01 to the right of the point. The line predicts 1.02 and the true tangent value is 1.020204, so the line is within two ten-thousandths - exactly what a tangent should do.
Section
Part 3
Intuition
A bank balance earning interest, a bacterial colony, a radioactive sample decaying: in every case, how fast the quantity changes depends on how much of it there is.
Twice as much money earns twice as much interest per year. Twice as many bacteria produce twice as many offspring per hour.
That sentence is a statement about a derivative. It says the rate of change is a constant multiple of the current value.
\[ \frac{dy}{dx} = k\,y \]
Exponential functions are exactly the functions that behave this way, and that is why their derivatives look the way they do.
Concept
Push a general exponential through the limit definition and something remarkable falls out.
\[ \frac{d}{dx}\left[b^{x}\right] = \lim_{h\to 0}\frac{b^{x+h}-b^{x}}{h} = b^{x}\cdot\lim_{h\to 0}\frac{b^{h}-1}{h} \]
The original function factors right back out. Whatever that leftover limit turns out to be, it is a constant - it has no variable in it at all.
So the derivative of an exponential is always a constant multiple of itself. The only question is which constant.
| base b | the limit at a step of 0.001 | its exact value |
|---|---|---|
| 2 | 0.693387 | 0.693147 |
| 2.5 | 0.916711 | 0.916291 |
| e = 2.718281828 | 1.000500 | 1.000000 |
| 3 | 1.099216 | 1.098612 |
One base makes that constant exactly one. That base is the number called e, and that is the whole reason it is worth a letter of its own.
Comparison
Comparison matrix
From Every exponential is its own derivative, up to a factor: refill the its exact value column from what you know. The rest of the table is as it appeared.
| base b | the limit at a step of 0.001 | its exact value |
|---|---|---|
| 2 | 0.693387 | 0.693147 |
| 2.5 | 0.916711 | 0.916291 |
| e = 2.718281828 | 1.000500 | 1.000000 |
| 3 | 1.099216 | 1.098612 |
Picture it
Animation
Shows: Every exponential is its own derivative, up to a factor — a rendered Manim animation.
Rendered with Manim.
Takeaway: Change the base and only a constant multiple appears.
Concept
the number e — The unique base whose exponential function has slope exactly 1 at the origin. Its value is about 2.718281828.
Because that leftover limit equals one for this base, the derivative is the function itself.
\[ \frac{d}{dx}\left[e^{x}\right] = e^{x} \]
This is the only function, apart from the constant zero and its own constant multiples, that comes back completely unchanged when you differentiate it.
Differentiate it a hundred times and nothing happens. That is why it shows up everywhere in science: it is the fixed point of the derivative.
Definition probe
Sort into buckets
Every line below is part of the definition of radian measure or of the number e — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: An exponential with any base — a rendered Manim animation.
Rendered with Manim.
Takeaway: The stray ln b is where base e earns its place.
Picture it
Animation
Shows: A tangent line on the exponential curve where slope equals height.
Height and slope, the same number everywhere.
Takeaway: At every point on this curve the height and the slope are the same number. That property is what singles out e among all the bases.
Picture it
Figure (svg): The natural exponential curve with tangent lines drawn where the height is one and where the height is about 2.72, each tangent having a slope equal to that height.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Pick any point on this curve, measure how high it is, and that number is also how steep it is right there.
Intuition
Figure (svg): The natural exponential curve with tangent lines drawn where the height is one and where the height is about 2.72, each tangent having a slope equal to that height.
Pick any point on this curve, measure how high it is, and that number is also how steep it is right there.
Where the curve is one unit tall, the tangent rises one unit for every unit across. Where it is about 2.72 tall, the tangent is about 2.72 times as steep.
That is the growth-proportional-to-size idea from a moment ago, drawn as a picture. The bigger it gets, the faster it grows, in exact proportion.
Worked example
Find the tangent line where the input is one.
\[ y = e^{x} \quad \text{at} \quad x = 1 \]
Find the point
Why: The height at that input is the number e itself, about 2.71828.
\[ y(1) = e^{1} = e \approx 2.71828 \]
Find the slope
Why: The derivative is the same function, so the slope at that input is also e. Height equals slope, exactly as the picture showed.
\[ y' = e^{x}, \qquad y'(1) = e \]
Write and simplify the line
Why: Point-slope form, then distribute. The e terms cancel and the tangent turns out to pass straight through the origin.
\[ y - e = e(x-1) \;\Longrightarrow\; y = ex \]
Verify the line hugs the curve
Why: Step 0.01 to the right. The curve is at 2.745601 and the line predicts 2.745465, a gap of about one ten-thousandth - the signature of a genuine tangent.
\[ e^{1.01} = 2.745601 \quad \text{vs.} \quad e(1.01) = 2.745465 \]
Picture it
Animation
Shows: Each line of the worked example "tangent to the natural exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Step 0.01 to the right. The curve is at 2.745601 and the line predicts 2.745465, a gap of about one ten-thousandth - the signature of a genuine tangent.
Concept
For any positive base, the leftover constant from the limit is the natural logarithm of that base.
\[ \frac{d}{dx}\left[b^{x}\right] = b^{x}\ln b \]
Check the table from before against this claim: the constant for base two was 0.693147, and that is exactly the natural log of two.
\[ \ln 2 = 0.693147, \qquad \ln 3 = 1.098612, \qquad \ln e = 1 \]
The natural exponential is not a special case with a different rule. It is the case where the extra factor happens to be one, so you never see it.
Estimation
Predict first
Nothing needs to be memorized here. Rewrite any base as a power of e and the chain rule produces the factor for you.
Commit before you compute: what does Worked example: where the natural-log factor comes from come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with base two at the origin
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The rule predicts a slope of the natural log of two, or 0.693147.
Worked example
Nothing needs to be memorized here. Rewrite any base as a power of e and the chain rule produces the factor for you.
\[ b = e^{\ln b} \quad \Longrightarrow \quad b^{x} = \left(e^{\ln b}\right)^{x} = e^{(\ln b)x} \]
Identify the outer and inner functions
Why: The outer function is the natural exponential; the inner function is the natural log of b times x, which is just a constant times x.
\[ \text{outer: } e^{u}, \qquad \text{inner: } u = (\ln b)\,x \]
Apply the chain rule
Why: The natural exponential differentiates to itself, then you multiply by the derivative of the inside, which is the constant natural log of b.
\[ \frac{d}{dx}\left[e^{(\ln b)x}\right] = e^{(\ln b)x}\cdot \ln b \]
Translate back to the original base
Why: The exponential of the natural log of b times x is the thing you started with, so the rule reads cleanly.
\[ = b^{x}\ln b \]
Verify with base two at the origin
Why: The rule predicts a slope of the natural log of two, or 0.693147. A symmetric difference quotient with a step of 0.001 gives 0.693267, and setting b equal to e returns the earlier rule because the natural log of e is one.
\[ \frac{2^{0.001}-2^{-0.001}}{0.002} = \frac{1.000693387-0.999306853}{0.002} = 0.693267 \]
Picture it
Animation
Shows: Each line of the worked example "where the natural-log factor comes from", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The rule predicts a slope of the natural log of two, or 0.693147. A symmetric difference quotient with a step of 0.001 gives 0.693267, and setting b equal to e returns the earlier rule because the natural log of e is one.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Seeing a base and an exponent and reaching for the power rule out of habit.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It predicts a slope of three times four, which is twelve.
The variable is upstairs, so this is an exponential, not a power. The power rule needs the variable in the base.
Why: It predicts a slope of three times four, which is twelve.
Trap
Seeing a base and an exponent and reaching for the power rule out of habit.
\[ \frac{d}{dx}\left[2^{x}\right] \stackrel{?}{=} x\cdot 2^{x-1} \]
Evaluate the claim at three
Why: It predicts a slope of three times four, which is twelve.
\[ \text{claim: } 3\cdot 2^{2} = 12 \]
A softer version of the same error
Why: Some students instead write the derivative of 2 to the x as just 2 to the x, copying the natural exponential rule. That predicts a slope of 8 - still wrong, because the natural-log factor is missing.
The variable is upstairs, so this is an exponential, not a power. The power rule needs the variable in the base.
\[ \frac{d}{dx}\left[2^{x}\right] = 2^{x}\ln 2 \]
Evaluate the rule at three
Why: Eight times the natural log of two is about 5.5452 - less than half of the claimed twelve.
\[ 2^{3}\ln 2 = 8(0.693147) = 5.545177 \]
Settle it numerically
Why: A symmetric difference quotient with a step of 0.001 gives about 5.546, right next to 5.545 and nowhere near 12 or 8.
\[ \frac{2^{3.001}-2^{2.999}}{0.002} = \frac{8.0055471-7.9944548}{0.002} = 5.5461 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
Eight times the natural log of two is about 5.5452 - less than half of the claimed twelve.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
It predicts a slope of three times four, which is twelve.
Ranking
Put in order
Put the moves of Worked example: an exponential with an inside function into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The outer function is the natural exponential; the inner function is the squaring.
Worked example
\[ y = e^{x^{2}} \]
Name the layers before touching anything
Why: The outer function is the natural exponential; the inner function is the squaring. Getting these backwards is the most common chain-rule failure.
\[ \text{outer: } e^{u}, \qquad \text{inner: } u = x^{2} \]
Differentiate the outer layer, leaving the inside alone
Why: The natural exponential returns itself, so this layer contributes an unchanged copy of the original expression.
\[ \frac{d}{dx}\left[e^{u}\right] = e^{u} = e^{x^{2}} \]
Multiply by the derivative of the inside
Why: The chain rule is a multiplication, not a substitution. The inside is x squared, whose derivative is two x.
\[ y' = e^{x^{2}}\cdot 2x = 2x\,e^{x^{2}} \]
Verify numerically at one
Why: The formula predicts two times e, or 5.436564. A symmetric difference quotient with a step of 0.001 gives 5.436573, so the extra factor of two x really does belong there.
\[ y'(1) = 2e \approx 5.436564, \qquad \frac{e^{(1.001)^{2}}-e^{(0.999)^{2}}}{0.002} = 5.436573 \]
Picture it
Animation
Shows: Each line of the worked example "an exponential with an inside function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts two times e, or 5.436564. A symmetric difference quotient with a step of 0.001 gives 5.436573, so the extra factor of two x really does belong there.
Prediction
Predict first
What is the derivative of 5 raised to the power x?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 5 to the x, times the natural log of 5
Why: Rewriting 5 to the x as e raised to the natural log of 5 times x and applying the chain rule leaves a factor of the natural log of 5, about 1.6094. So the derivative is 5 to the x times the natural log of 5.
Check
The variable sits in the exponent here. Choose the rule that matches that, not the one that matches the shape.
\[ g(x) = 5^{x} \]
Check your understanding
What is the derivative of 5 raised to the power x?
Answer: A
Why: Rewriting 5 to the x as e raised to the natural log of 5 times x and applying the chain rule leaves a factor of the natural log of 5, about 1.6094. So the derivative is 5 to the x times the natural log of 5.
Section
Part 4
Picture it
Figure (svg): The natural logarithm curve rising steeply near the vertical asymptote and flattening out to the right, with two tangent segments showing a steep slope at one and a gentler slope at three.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The exponential curve gets steeper and steeper as you move right. Reflect it across the diagonal line to get the logarithm, and steepness turns into flatness.
Intuition
Figure (svg): The natural logarithm curve rising steeply near the vertical asymptote and flattening out to the right, with two tangent segments showing a steep slope at one and a gentler slope at three.
The exponential curve gets steeper and steeper as you move right. Reflect it across the diagonal line to get the logarithm, and steepness turns into flatness.
Where the input is one the log climbs at a slope of one. By the time the input is three it has slowed to a third of that. At one hundred it barely moves at all.
So the slope is shrinking in exact proportion to how far right you are. That already tells you the derivative must be one over the input.
Picture it
Animation
Shows: A log is an exponential read backwards — a rendered Manim animation.
Rendered with Manim.
Takeaway: The log rule falls straight out of the exponential rule.
Concept
\[ \frac{d}{dx}\left[\ln x\right] = \frac{1}{x}, \qquad x > 0 \]
This is the rule that fills the one hole in the power rule for antiderivatives, and it is worth noticing how strange it is: a transcendental function whose derivative is a plain algebraic fraction.
The domain restriction is real. The natural log is only defined for positive inputs, so the derivative statement only claims anything there.
Picture it
Animation
Shows: The natural logarithm rising steeply then flattening.
Steep near zero, nearly flat far out.
Takeaway: Its slope is one over x — enormous close to zero and vanishing to the right, which is exactly what the graph is doing.
Hypothesis
Predict first
Worked example: deriving the natural-log rule is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Differentiate both sides with respect to x
Why: The left side has y inside it and y depends on x, so the chain rule leaves a derivative factor behind. The right side is just x, whose derivative is one.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Name the log, then undo it with an exponential. That converts an unfamiliar derivative into a familiar one.
\[ y = \ln x \quad \Longleftrightarrow \quad e^{y} = x \]
Differentiate both sides with respect to x
Why: The left side has y inside it and y depends on x, so the chain rule leaves a derivative factor behind. The right side is just x, whose derivative is one.
\[ e^{y}\cdot\frac{dy}{dx} = 1 \]
Solve for the derivative
Why: Divide by the exponential factor, which is never zero, so the division is always legal.
\[ \frac{dy}{dx} = \frac{1}{e^{y}} \]
Replace the exponential with x
Why: The starting relation says that exponential IS x, so the answer comes out in terms of x, which is what a derivative must be.
\[ \frac{dy}{dx} = \frac{1}{x} \]
Verify at one and at e
Why: The rule predicts a slope of one at an input of one, and a symmetric difference quotient with a step of 0.001 gives 1.0000003. It also predicts about 0.3679 at an input of e, matching the visibly gentler tangent in the picture.
\[ \frac{\ln(1.001)-\ln(0.999)}{0.002} = \frac{0.0009995003+0.0010005003}{0.002} = 1.0000003 \]
Picture it
Animation
Shows: Each line of the worked example "deriving the natural-log rule", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The rule predicts a slope of one at an input of one, and a symmetric difference quotient with a step of 0.001 gives 1.0000003. It also predicts about 0.3679 at an input of e, matching the visibly gentler tangent in the picture.
Concept
Almost no exam problem asks for the log of a bare variable. There is usually something inside.
\[ \frac{d}{dx}\left[\ln u\right] = \frac{u'}{u} \]
Say it out loud as derivative of the inside, over the inside. That phrase covers every natural-log derivative you will ever write.
\[ \frac{d}{dx}\left[\ln(\sin x)\right] = \frac{\cos x}{\sin x} = \cot x \]
Worked example
\[ y = \ln\!\left(x^{2}+1\right) \]
Write down the inside and its derivative first
Why: Doing this before applying the rule stops you from forgetting the numerator, which is the usual mistake.
\[ u = x^{2}+1, \qquad u' = 2x \]
Assemble derivative of the inside over the inside
Why: The outer log contributes the reciprocal, and the chain rule multiplies by the inner derivative, which lands in the numerator.
\[ y' = \frac{2x}{x^{2}+1} \]
Notice the domain is everything
Why: The inside is never zero or negative, so unlike a plain natural log this function and its derivative are defined for every real input.
Verify numerically at one
Why: The formula predicts a slope of two over two, which is one. A symmetric difference quotient with a step of 0.001 gives 1.0000, so the numerator factor of two x is genuinely there.
\[ \frac{\ln(2.002001)-\ln(1.998001)}{0.002} = \frac{0.6941472-0.6921472}{0.002} = 1.0000 \]
Concept
Change of base turns any logarithm into a constant multiple of the natural log.
\[ \log_{b} x = \frac{\ln x}{\ln b} \]
Differentiate the constant multiple
Why: The natural log of b is a number, so it just sits in the denominator while the natural log differentiates normally.
\[ \frac{d}{dx}\left[\log_{b} x\right] = \frac{1}{x\ln b} \]
Notice where the natural-log factor goes. For an exponential it multiplies; for a logarithm it divides. Getting those two backwards is a favorite exam trap.
| function | derivative | the log factor |
|---|---|---|
| b to the x | b to the x times ln b | multiplies |
| log base b of x | 1 over x times ln b | divides |
Picture it
Animation
Shows: A logarithm with any base — a rendered Manim animation.
Rendered with Manim.
Takeaway: Change of base first, then the rule you already know.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Carrying the exponential's headline property over to the logarithm.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The natural log of one is zero, so this claim says the log curve is perfectly flat there.
Only the natural exponential is its own derivative. Its inverse behaves completely differently.
Why: The natural log of one is zero, so this claim says the log curve is perfectly flat there.
Trap
Carrying the exponential's headline property over to the logarithm.
\[ \frac{d}{dx}\left[\ln x\right] \stackrel{?}{=} \ln x \]
Test the claim at an input of one
Why: The natural log of one is zero, so this claim says the log curve is perfectly flat there.
\[ \text{claim: } \left.\frac{d}{dx}\ln x\right|_{x=1} = \ln 1 = 0 \]
Look at the curve
Why: The log is climbing briskly through that point - it is negative just to the left and positive just to the right. A flat tangent is impossible there.
Only the natural exponential is its own derivative. Its inverse behaves completely differently.
\[ \frac{d}{dx}\left[\ln x\right] = \frac{1}{x} \]
Test it at an input of one
Why: The rule predicts a slope of one, matching the brisk climb you can see in the picture.
\[ \left.\frac{1}{x}\right|_{x=1} = 1 \]
Confirm numerically
Why: The measured slope at that point is 1.0000003, so the reciprocal rule is right and the self-derivative claim is off by a full unit.
Notation
Annotate
From Trap: the log is not its own derivative — read this one piece at a time. What is each part doing?
On: \( \left.\frac{1}{x}\right|_{x=1} = 1 \)
Section
Part 5
Constraint
Discussion prompt
Run Playbook: which rule, in which order with this step confiscated:
If the last thing you would do is add or subtract, split the problem there and differentiate each piece separately.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
Pattern
The rules are easy. Choosing among them under time pressure is the actual skill. Work from the outside in.
The chain rule is the one that hides inside the others. Any time a product or quotient term has something inside it, that term still needs its own inner factor.
Edge cases
Discussion prompt
Playbook: which rule, in which order works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
The rules are easy. Choosing among them under time pressure is the actual skill. Work from the outside in.
Step zero
Discussion prompt
Worked example: a product where both factors are chained — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Identify the last operation as a multiplication
Answer:
Worked example
\[ y = e^{2x}\sin(3x) \]
Identify the last operation as a multiplication
Why: You would compute the exponential, compute the sine, and multiply. Multiplication last means product rule first.
\[ u = e^{2x}, \qquad v = \sin(3x) \]
Differentiate each factor with the chain rule
Why: Each factor has an inside function, so each one contributes its own inner factor. This is where most of the lost points live.
\[ u' = 2e^{2x}, \qquad v' = 3\cos(3x) \]
Assemble the product rule
Why: Derivative of the first times the second, plus the first times the derivative of the second.
\[ y' = 2e^{2x}\sin(3x) + 3e^{2x}\cos(3x) \]
Factor the common exponential
Why: The exponential appears in both terms and is never zero, so pulling it out is safe and makes the structure obvious.
\[ y' = e^{2x}\left[2\sin(3x) + 3\cos(3x)\right] \]
Verify numerically at the origin
Why: The formula predicts one times zero plus three, which is three. A symmetric difference quotient with a step of 0.001 gives 3.000003, so both inner factors are in the right places.
\[ y'(0) = 1\left[2(0)+3(1)\right] = 3, \qquad \frac{y(0.001)-y(-0.001)}{0.002} = 3.000003 \]
Picture it
Animation
Shows: Each line of the worked example "a product where both factors are chained", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts one times zero plus three, which is three. A symmetric difference quotient with a step of 0.001 gives 3.000003, so both inner factors are in the right places.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Running the product rule correctly, then differentiating the sine as if the inside were just the bare variable.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It predicts a slope of about negative 1.634.
The product rule tells you to differentiate the sine factor. Differentiating it correctly means using the chain rule on it.
Why: It predicts a slope of about negative 1.634.
Trap
Running the product rule correctly, then differentiating the sine as if the inside were just the bare variable.
\[ y = x^{2}\sin(5x) \;\Longrightarrow\; y' \stackrel{?}{=} 2x\sin(5x) + x^{2}\cos(5x) \]
Evaluate the claim at one
Why: It predicts a slope of about negative 1.634.
\[ 2\sin 5 + \cos 5 = -1.9178 + 0.2837 = -1.6342 \]
Notice what was dropped
Why: The second term needed the derivative of the inside, which is five. Missing it scales that whole term down by a factor of five.
The product rule tells you to differentiate the sine factor. Differentiating it correctly means using the chain rule on it.
\[ y' = 2x\sin(5x) + 5x^{2}\cos(5x) \]
Evaluate the rule at one
Why: The corrected version predicts about negative 0.4995 - a completely different number, not a small correction.
\[ 2\sin 5 + 5\cos 5 = -1.9178 + 1.4183 = -0.4995 \]
Settle it numerically
Why: A symmetric difference quotient with a step of 0.001 gives negative 0.49955, which matches the corrected answer and rules out the other one entirely.
\[ \frac{y(1.001)-y(0.999)}{0.002} = \frac{-0.9594100+0.9584109}{0.002} = -0.49955 \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Missing information
Discussion prompt
This function shows up whenever you compare growth rates. Find its derivative and its highest point.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The bottom is a single term, but it is a variable, so dividing through would not simplify anything here. The quotient rule is the right call.
Worked example
This function shows up whenever you compare growth rates. Find its derivative and its highest point.
\[ y = \frac{\ln x}{x}, \qquad x > 0 \]
Set up the quotient rule
Why: The bottom is a single term, but it is a variable, so dividing through would not simplify anything here. The quotient rule is the right call.
\[ \frac{d}{dx}\left[\frac{u}{v}\right] = \frac{vu'-uv'}{v^{2}}, \qquad u=\ln x,\; v=x \]
Substitute the two derivatives
Why: The natural log gives the reciprocal, and the bare variable gives one. Keep the order in the numerator: bottom times top-prime first.
\[ y' = \frac{x\cdot\frac{1}{x} - \ln x \cdot 1}{x^{2}} \]
Simplify the numerator
Why: The x cancels against its own reciprocal, leaving a clean one.
\[ y' = \frac{1-\ln x}{x^{2}} \]
Find where the tangent is horizontal
Why: A fraction is zero exactly when its numerator is zero, and the natural log equals one only at the input e.
\[ 1 - \ln x = 0 \;\Longrightarrow\; x = e \]
Verify the critical point is really the peak
Why: At the input e the function equals 0.3678794. Just to the left, at 2.5, it is 0.3665163; just to the right, at 3, it is 0.3662041. Both neighbors are lower, so the horizontal tangent really sits on a maximum.
\[ y(e) = \frac{1}{e} \approx 0.3678794 \]
Picture it
Animation
Shows: Each line of the worked example "a logarithm over a variable", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At the input e the function equals 0.3678794. Just to the left, at 2.5, it is 0.3665163; just to the right, at 3, it is 0.3662041. Both neighbors are lower, so the horizontal tangent really sits on a maximum.
Check
Two variable factors multiplied together, one of them a logarithm. Work it out before choosing.
\[ h(x) = x^{3}\ln x \]
Check your understanding
What is the derivative of x cubed times the natural log of x?
Answer: A
Why: The product rule gives 3x squared times ln x, plus x cubed times the derivative of the log. That second piece is x cubed times one over x, which simplifies to x squared. So the answer is 3x squared ln x plus x squared.
Estimation
Predict first
This one looks brutal and collapses to a single term. It is worth seeing once.
Commit before you compute: what does Worked example: a famously ugly log that simplifies come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify numerically at the origin
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The answer predicts the secant of zero, which is one.
Worked example
This one looks brutal and collapses to a single term. It is worth seeing once.
\[ y = \ln\!\left(\sec x + \tan x\right) \]
Use the derivative-of-the-inside-over-the-inside form
Why: The outer function is a natural log, so the whole answer is one fraction with the inside on the bottom.
\[ y' = \frac{\frac{d}{dx}\left[\sec x + \tan x\right]}{\sec x + \tan x} \]
Differentiate the inside term by term
Why: Secant gives secant times tangent, and tangent gives secant squared. Both rules were rebuilt earlier from the quotient rule.
\[ y' = \frac{\sec x\tan x + \sec^{2} x}{\sec x + \tan x} \]
Factor a secant out of the numerator
Why: Both numerator terms contain a secant, and what is left inside the bracket is exactly the denominator.
\[ y' = \frac{\sec x\left(\tan x + \sec x\right)}{\sec x + \tan x} = \sec x \]
Verify numerically at the origin
Why: The answer predicts the secant of zero, which is one. A symmetric difference quotient with a step of 0.001 gives 1.0000, confirming the cancellation was legitimate.
\[ \frac{\ln(1.0010005)-\ln(0.9990005)}{0.002} = \frac{0.0010000+0.0010000}{0.002} = 1.0000 \]
Picture it
Animation
Shows: Each line of the worked example "a famously ugly log that simplifies", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The answer predicts the secant of zero, which is one. A symmetric difference quotient with a step of 0.001 gives 1.0000, confirming the cancellation was legitimate.
Commit first
Predict first
What is the derivative of e raised to the power sine x?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: cosine x times e to the power sine x
Why: The outer function is the natural exponential, which returns itself unchanged, and the chain rule then multiplies by the derivative of the exponent. That derivative is cosine x, so the answer is cosine x times e to the sine x.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
One function is sitting inside another here. Name the outer layer before you differentiate.
\[ p(x) = e^{\sin x} \]
Check your understanding
What is the derivative of e raised to the power sine x?
Answer: A
Why: The outer function is the natural exponential, which returns itself unchanged, and the chain rule then multiplies by the derivative of the exponent. That derivative is cosine x, so the answer is cosine x times e to the sine x.
Worked example
\[ y = x^{2}\,3^{x} \]
Separate the power from the exponential
Why: The first factor has the variable in the base, so it is a power. The second has the variable upstairs, so it is an exponential. Different rules, one for each.
Apply the product rule
Why: Derivative of the first times the second, plus the first times the derivative of the second.
\[ y' = 2x\cdot 3^{x} + x^{2}\cdot\frac{d}{dx}\left[3^{x}\right] \]
Fill in the general exponential rule
Why: The base is not e, so the natural log of three comes along as a multiplying factor.
\[ y' = 2x\,3^{x} + x^{2}\,3^{x}\ln 3 \]
Factor for the clean form
Why: Both terms share an x and a power of three, so the answer compresses neatly.
\[ y' = x\,3^{x}\left(2 + x\ln 3\right) \]
Verify numerically at one
Why: The formula predicts three times the quantity two plus 1.098612, which is 9.29584. A symmetric difference quotient with a step of 0.001 gives 9.29582, so the natural-log factor is exactly where it belongs.
\[ y'(1) = 3\left(2+\ln 3\right) = 3(3.098612) = 9.29584 \]
Picture it
Animation
Shows: Each line of the worked example "a product with a general base", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts three times the quantity two plus 1.098612, which is 9.29584. A symmetric difference quotient with a step of 0.001 gives 9.29582, so the natural-log factor is exactly where it belongs.
Prediction
Predict first
What is the derivative of the sine of x squared?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: 2x times the cosine of x squared
Why: The outer function is sine, which becomes cosine while the inside is left untouched, and the chain rule then multiplies by the derivative of the inside, which is 2x. The result is 2x times the cosine of x squared.
Check
Two layers. Decide which one is outer, then differentiate.
\[ q(x) = \sin\!\left(x^{2}\right) \]
Check your understanding
What is the derivative of the sine of x squared?
Answer: A
Why: The outer function is sine, which becomes cosine while the inside is left untouched, and the chain rule then multiplies by the derivative of the inside, which is 2x. The result is 2x times the cosine of x squared.
Section
Part 6
Concept
The inverse trig functions undo sine, cosine, and tangent. Their derivatives are startling: no trig appears in the answers at all.
\[ \frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^{2}}}, \qquad -1 < x < 1 \]
The domain restriction is doing real work. As the input approaches either endpoint the denominator collapses toward zero and the slope blows up.
That matches the picture: the arcsine graph turns vertical at both ends, because it is the sine graph reflected, and sine is flat at its peak and trough.
Explain it
Discussion prompt
Explain Rule seven: arcsine to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The inverse trig functions undo sine, cosine, and tangent. Their derivatives are startling: no trig appears in the answers at all.
Picture it
Animation
Shows: Rule seven: arcsine — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every inverse-trig rule comes from the same implicit trick.
Intuition
Nothing here is memorized. Name the inverse, undo it, and differentiate what is left.
\[ y = \arcsin x \quad \Longleftrightarrow \quad \sin y = x \]
Differentiate both sides. The left side has y inside it, so the chain rule leaves a derivative factor behind, exactly as it did for the natural log.
\[ \cos y \cdot \frac{dy}{dx} = 1 \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{1}{\cos y} \]
Now trade the cosine for something in terms of x, using the Pythagorean identity and the fact that arcsine only outputs angles where cosine is not negative.
\[ \cos y = \sqrt{1-\sin^{2} y} = \sqrt{1-x^{2}} \]
That is the whole story. The square root is the leg of a right triangle whose hypotenuse is one and whose opposite side is the input.
Analogy
Discussion prompt
Explain Where that square root comes from by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Nothing here is memorized. Name the inverse, undo it, and differentiate what is left.
Concept
\[ \frac{d}{dx}\left[\arccos x\right] = -\frac{1}{\sqrt{1-x^{2}}}, \qquad \frac{d}{dx}\left[\arctan x\right] = \frac{1}{1+x^{2}} \]
The co-function minus sign shows up one more time. Arccosine's derivative is the exact negative of arcsine's - the two curves add up to a constant, so their slopes must cancel.
Arctangent is the friendliest of the three: its denominator is never zero, so it is differentiable everywhere, and its slope quietly dies off as the input grows.
| function | derivative | where it is valid |
|---|---|---|
| arcsine | 1 over the square root of 1 minus x squared | strictly between -1 and 1 |
| arccosine | negative 1 over the same square root | strictly between -1 and 1 |
| arctangent | 1 over 1 plus x squared | every real number |
Comparison
Comparison matrix
From Rules eight and nine: arccosine and arctangent: refill the where it is valid column from what you know. The rest of the table is as it appeared.
| function | derivative | where it is valid |
|---|---|---|
| arcsine | 1 over the square root of 1 minus x squared | strictly between -1 and 1 |
| arccosine | negative 1 over the same square root | strictly between -1 and 1 |
| arctangent | 1 over 1 plus x squared | every real number |
Step zero
Discussion prompt
Worked example: arctangent with an inside function — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Write the outer rule with the inside held as a block
Answer:
Worked example
\[ y = \arctan\!\left(x^{2}\right) \]
Write the outer rule with the inside held as a block
Why: The rule says one over one plus the input squared. Here the input is the whole inside function, so it gets squared as a unit.
\[ \frac{d}{dx}\left[\arctan u\right] = \frac{1}{1+u^{2}}\cdot u', \qquad u = x^{2} \]
Substitute the inside and square it correctly
Why: Squaring x squared gives x to the fourth, not two x squared. This is where careless answers go wrong.
\[ \frac{1}{1+\left(x^{2}\right)^{2}} = \frac{1}{1+x^{4}} \]
Multiply by the derivative of the inside
Why: The chain rule factor is the derivative of x squared, which is two x, and it goes in the numerator.
\[ y' = \frac{2x}{1+x^{4}} \]
Verify numerically at one
Why: The formula predicts two over two, which is one. A symmetric difference quotient with a step of 0.001 gives 1.0000, so both the fourth power and the inner factor are right.
\[ y'(1) = \frac{2}{1+1} = 1, \qquad \frac{\arctan(1.002001)-\arctan(0.998001)}{0.002} = 1.0000 \]
Picture it
Animation
Shows: Each line of the worked example "arctangent with an inside function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts two over two, which is one. A symmetric difference quotient with a step of 0.001 gives 1.0000, so both the fourth power and the inner factor are right.
Elimination
Eliminate the wrong options
What is the derivative of the arctangent of 3x?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Substituting the inside into the arctangent rule gives one over one plus the square of 3x, and squaring 3x gives 9x squared. The chain rule then multiplies by the derivative of the inside, which is 3, putting a 3 in the numerator.
Check
Watch both places the inside function has to appear.
\[ r(x) = \arctan(3x) \]
Check your understanding
What is the derivative of the arctangent of 3x?
Answer: A
Why: Substituting the inside into the arctangent rule gives one over one plus the square of 3x, and squaring 3x gives 9x squared. The chain rule then multiplies by the derivative of the inside, which is 3, putting a 3 in the numerator.
Concept
Pull a mass hanging on a spring down four centimeters and let go. Ignoring friction, its position over time is a cosine wave.
\[ s(t) = 4\cos(3t) \quad \text{centimeters, } t \text{ in seconds} \]
The four is the amplitude, the farthest it ever gets from the resting point. The three controls how fast it oscillates.
\[ \text{period} = \frac{2\pi}{3} \approx 2.094 \text{ seconds} \]
Velocity is the derivative of position and acceleration is the derivative of velocity, so the trig rules give you the entire motion from one formula.
Counterexample
Discussion prompt
Pull a mass hanging on a spring down four centimeters and let go. Ignoring friction, its position over time is a cosine wave.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The four is the amplitude, the farthest it ever gets from the resting point. The three controls how fast it oscillates.
Ranking
Put in order
Put the moves of Worked example: velocity and acceleration of the spring into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Cosine gives negative sine, and the chain rule multiplies by the derivative of the inside, which is three.
Worked example
\[ s(t) = 4\cos(3t) \]
Differentiate once for velocity
Why: Cosine gives negative sine, and the chain rule multiplies by the derivative of the inside, which is three. The four rides along as a constant factor.
\[ v(t) = s'(t) = -12\sin(3t) \quad \text{centimeters per second} \]
Differentiate again for acceleration
Why: Sine gives cosine, the inside contributes another three, and the leading minus sign stays put.
\[ a(t) = v'(t) = -36\cos(3t) \quad \text{centimeters per second squared} \]
Read the motion off the three formulas
Why: At release the mass is at its extreme, momentarily at rest, and pulled hardest back toward the middle. As it passes the middle it is moving fastest and feels no force at all.
| time (s) | position (cm) | velocity (cm/s) | acceleration (cm per s squared) |
|---|---|---|---|
| 0 | 4 | 0 | -36 |
| 0.5236 | 0 | -12 | 0 |
| 1.0472 | -4 | 0 | 36 |
| 1.5708 | 0 | 12 | 0 |
Verify that acceleration is proportional to displacement
Why: Negative nine times the position formula reproduces the acceleration formula exactly, and the table confirms it at every listed time. That proportionality, with the minus sign, is the definition of simple harmonic motion and is Hooke's law in disguise.
\[ -9\,s(t) = -9\left[4\cos(3t)\right] = -36\cos(3t) = a(t) \]
Picture it
Animation
Shows: Each line of the worked example "velocity and acceleration of the spring", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Negative nine times the position formula reproduces the acceleration formula exactly, and the table confirms it at every listed time. That proportionality, with the minus sign, is the definition of simple harmonic motion and is Hooke's law in disguise.
Ranking
Put in order
These are the steps of The complete transcendental toolkit, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Nine rules. Everything else in this deck was these nine plus the chain, product, and quotient rules.
| function | derivative |
|---|---|
| sine | cosine |
| cosine | negative sine |
| tangent | secant squared |
| secant | secant times tangent |
| natural exponential | itself |
| b to the x | b to the x times ln b |
| natural log of x | 1 over x |
| log base b of x | 1 over x times ln b |
| arctangent | 1 over 1 plus x squared |
Add a co to any trig name and attach a minus sign to get the other three trig rules. That is the whole table.
Real world
Discussion prompt
Outside this lesson: where does Derivatives of Trig, Exponential, and Log Functions actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of The complete transcendental toolkit is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck assembles the full transcendental toolkit. It derives sine and cosine from the two special limits, rebuilds the other four trigonometric functions with the quotient rule, and covers the natural exponential that is its own derivative, general bases with their natural-log factor, the natural and general logarithm, and the inverse trigonometric derivatives - then combines all of it with the chain, product, and quotient rules. It targets the sign error on cosine and the co-functions, the degrees-versus-radians error, the missing natural-log factor on a general exponential, and the confusion between the logarithm and the exponential rules.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Sine and Cosine · The Other Four Trig Functions · The Exponential Functions · Logarithms · Putting Them Together · Inverse Trig and Motion. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now differentiate essentially any function a first-semester course will show you, and you can rebuild any rule you forget.
| if you see | reach for |
|---|---|
| something inside a function | the chain rule, and do not lose the inner factor |
| a base that is not e | the natural log of that base |
| a co-function | a minus sign in the derivative |
| degrees | convert to radians before differentiating |
Next up: implicit and logarithmic differentiation, where these same rules let you differentiate equations that were never solved for y in the first place.
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