This deck covers the shortcut rules that replace the limit definition: the constant, power, constant-multiple, and sum and difference rules, plus the natural exponential. It also covers rewriting radicals and reciprocals as powers before differentiating, higher-order derivatives, tangent lines, and horizontal tangents. It targets the classic errors: using the power rule on a constant base, forgetting to rewrite, assuming the derivative distributes over a product, and slipping a sign on a negative exponent.
Subject: Calculus I · 146 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 06
Constant, power, constant-multiple, sum and difference - and how to rewrite an expression so the rules apply.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Power, Constant, Sum, and Difference Rules: without looking back, what was the main idea of The Derivative: Definition, Meaning, and Differentiability, 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 builds the derivative from average rates of change and secant slopes up to the limit definition. It then computes derivatives from that definition for linear, quadratic, rational, and square-root functions, reads the derivative off a graph and in real units, and settles when differentiability fails. It targets the dropped minus sign in the difference quotient, cancelling h before it is a factor, the belief that continuous means differentiable, and confusing a function value with a slope.
Section
Section 1
Concept
You already know what a derivative is: the limit of secant slopes.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
It works for every function. It also takes half a page of algebra for even a simple one.
This deck builds a small set of shortcut rules so the limit work gets done once, in general, and never again.
Counterexample
Discussion prompt
You already know what a derivative is: the limit of secant slopes.
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.
Picture it
Animation
Shows: Correct, but unusable at scale — a rendered Manim animation.
Rendered with Manim.
Takeaway: The rules exist so the limit is computed once and never again.
Intuition
You could compute seven times eight by adding eight to itself seven times. You did that once in third grade, wrote the answer down, and now you just recall it.
Every rule in this deck was proved once from the limit definition. After that, differentiating becomes recall plus a little arithmetic.
That is the whole trade: a few minutes of proof now buys you speed for the rest of the course.
Analogy
Discussion prompt
Explain Do the hard work once, then recall it 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:
You could compute seven times eight by adding eight to itself seven times. You did that once in third grade, wrote the answer down, and now you just recall it.
Ranking
Put in order
Put the moves of Worked example: the long way, one last time 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. This is the definition; nothing is assumed yet.
Worked example
Differentiate this from the definition, so you can watch the shortcut appear.
\[ f(x) = x^2 \]
Write the difference quotient
Why: This is the definition; nothing is assumed yet.
\[ \frac{f(x+h) - f(x)}{h} = \frac{(x+h)^2 - x^2}{h} \]
Expand the numerator and cancel
Why: Squaring the binomial produces a copy of x squared that the subtraction removes.
\[ (x+h)^2 - x^2 = x^2 + 2xh + h^2 - x^2 = 2xh + h^2 \]
Factor h out and divide
Why: Every surviving term has an h in it, which is exactly why the quotient stays finite.
\[ \frac{2xh + h^2}{h} = 2x + h \]
Take the limit
Why: Now h can safely go to zero, because there is no longer an h in a denominator.
\[ f'(x) = \lim_{h \to 0} (2x + h) = 2x \]
Verify numerically at the point where x equals 3
Why: The formula predicts a slope of 6. Real secant slopes computed from the function march straight to 6.
| h | secant slope at x = 3 |
|---|---|
| 0.1 | 6.1 |
| 0.01 | 6.01 |
| 0.001 | 6.001 |
Concept
The simplest rule of all: a constant function never changes, so its rate of change is zero.
\[ \frac{d}{dx}\left[c\right] = 0 \qquad \text{for any constant } c \]
constant rule — The derivative of a constant function is zero everywhere. The graph is a horizontal line, and a horizontal line has slope zero at every point.
Picture it
Animation
Shows: A flat road has no slope — a rendered Manim animation.
Rendered with Manim.
Takeaway: Nothing changes, so the rate of change is zero everywhere.
Picture it
Figure (svg): A horizontal line drawn above the x-axis on a pair of axes; its height is the same at every point.
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:
If your car's odometer reads the same number all afternoon, your speed was zero all afternoon. There is nothing to compute.
Intuition
Figure (svg): A horizontal line drawn above the x-axis on a pair of axes; its height is the same at every point.
If your car's odometer reads the same number all afternoon, your speed was zero all afternoon. There is nothing to compute.
Constants are the background noise of a derivative problem: they matter for the function's value and not at all for its slope.
Explain it
Discussion prompt
Explain A flat road has no slope 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:
If your car's odometer reads the same number all afternoon, your speed was zero all afternoon. There is nothing to compute.
Step zero
Discussion prompt
Worked example: the constant rule from the definition — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Build the difference quotient
Answer:
Worked example
It only takes two lines, and it is worth seeing once.
\[ f(x) = 7 \]
Build the difference quotient
Why: The function outputs 7 no matter what goes in, so both pieces of the numerator are 7.
\[ \frac{f(x+h) - f(x)}{h} = \frac{7 - 7}{h} = \frac{0}{h} = 0 \]
Note that the division is legal
Why: In a limit as h goes to zero, h is never actually zero, so dividing by h is allowed the whole way.
Take the limit of the constant zero
Why: The quotient is the number zero for every nonzero h, so the limit is zero.
\[ f'(x) = \lim_{h \to 0} 0 = 0 \]
Verify against the graph
Why: The graph of the function is the horizontal line at height 7. Pick any two points on it: the rise is 0, so every secant slope is already 0 before any limit is taken.
| two points on the graph | rise | run | slope |
|---|---|---|---|
| (1, 7) and (5, 7) | 0 | 4 | 0 |
| (2, 7) and (2.1, 7) | 0 | 0.1 | 0 |
Concept
This is the workhorse of the entire course.
\[ \frac{d}{dx}\left[x^{n}\right] = n\,x^{\,n-1} \]
In words: bring the exponent down in front, then knock the exponent down by one.
| function | bring down | subtract one | derivative |
|---|---|---|---|
| x to the 5th | 5 | 5 - 1 = 4 | 5 times x to the 4th |
| x to the 12th | 12 | 12 - 1 = 11 | 12 times x to the 11th |
| x squared | 2 | 2 - 1 = 1 | 2x |
Picture it
Animation
Shows: The power rule with two worked instances underneath.
Verify it once from the definition. Then trust it.
Takeaway: The exponent drops in front and steps down by one. Check it against the difference quotient once, and you never need to again.
Fill the middle
Fill in the blanks
From Where the exponent out front comes from — finish the line. Write what belongs on the right of the equals sign before you look.
\fracn\,x^{n-1} + (\text{terms still containing } h)___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The first term cancels with the function value, and every remaining term carries at least one h.
Intuition
Expand the binomial in the difference quotient and watch which term survives.
\[ (x+h)^n = x^n + n\,x^{n-1}h + (\text{terms with } h^2 \text{ or higher}) \]
Subtract the leading term, then divide by h
Why: The first term cancels with the function value, and every remaining term carries at least one h.
\[ \frac{(x+h)^n - x^n}{h} = n\,x^{n-1} + (\text{terms still containing } h) \]
Letting h go to zero kills everything that still has an h in it. Exactly one term is left standing, and it is the rule.
The exponent appears out front because there are that many ways to pick the single h out of the expansion. That is the whole secret.
Translation
\( \frac{(x+h)^n - x^n}{h} = n\,x^{n-1} + (\text{terms still containing } h) \)
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.
Picture it
Animation
Shows: A line of slopes sweeping as its coefficient changes.
Every input has its own slope.
Takeaway: Each input has its own slope, and collecting them all produces a new function — which is why the derivative is a machine, not a number.
Estimation
Predict first
Test the rule on a cube, where the algebra is still short enough to check.
Commit before you compute: what does Worked example: does the rule match the definition? 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 point where x equals 2
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 12.
Worked example
Test the rule on a cube, where the algebra is still short enough to check.
\[ f(x) = x^3 \]
Apply the power rule first
Why: Bring the 3 down, drop the exponent to 2. This is the prediction we are about to test.
\[ f'(x) = 3x^{2} \]
Now expand the difference quotient the long way
Why: Cubing the binomial gives four terms; the leading one cancels against the subtraction.
\[ \frac{(x+h)^3 - x^3}{h} = \frac{3x^2h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2 \]
Take the limit
Why: Both remaining terms carry an h, so both vanish. The definition agrees with the rule.
\[ \lim_{h \to 0}\left(3x^2 + 3xh + h^2\right) = 3x^2 \]
Verify numerically at the point where x equals 2
Why: The rule predicts a slope of 12. The secant slopes computed from the cube close in on 12.
| h | secant slope at x = 2 |
|---|---|
| 0.1 | 12.61 |
| 0.01 | 12.0601 |
| 0.001 | 12.006001 |
Picture it
Animation
Shows: Each line of the worked example "does the rule match the definition?", 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 12. The secant slopes computed from the cube close in on 12.
Concept
Exponent one. The graph is a line of slope one, and the rule agrees.
\[ \frac{d}{dx}\left[x\right] = \frac{d}{dx}\left[x^{1}\right] = 1 \cdot x^{0} = 1 \]
Exponent zero. Anything to the zero power is the constant one, so its derivative had better be zero.
\[ \frac{d}{dx}\left[x^{0}\right] = \frac{d}{dx}\left[1\right] = 0 \]
Both are just the constant rule and the power rule shaking hands. Nothing new to memorize.
Concept
The power rule is stated for whole numbers first because that proof is easy. It is in fact true for every real exponent.
\[ \frac{d}{dx}\left[x^{n}\right] = n\,x^{\,n-1} \quad \text{for every real number } n \]
| exponent type | example function | derivative |
|---|---|---|
| negative | x to the power negative 4 | negative 4 times x to the power negative 5 |
| fractional | x to the one half | one half times x to the power negative one half |
| decimal | x to the 2.5 | 2.5 times x to the 1.5 |
That single fact is why the next section is about rewriting: if you can turn an expression into a power, you can differentiate it.
Comparison
Comparison matrix
From The exponent does not have to be a whole number: refill the example function column from what you know. The rest of the table is as it appeared.
| exponent type | example function | derivative |
|---|---|---|
| negative | x to the power negative 4 | negative 4 times x to the power negative 5 |
| fractional | x to the one half | one half times x to the power negative one half |
| decimal | x to the 2.5 | 2.5 times x to the 1.5 |
Picture it
Animation
Shows: The exponent does not have to be a whole number — a rendered Manim animation.
Rendered with Manim.
Takeaway: Rewrite into a power and the power rule reaches everything.
Missing information
Discussion prompt
Same recipe, and the only new work is arithmetic with a negative number.
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 exponent is negative four. Read it as a single signed number, not as a minus sign floating next to a 4.
Worked example
Same recipe, and the only new work is arithmetic with a negative number.
\[ f(x) = x^{-4} \]
Identify the exponent
Why: The exponent is negative four. Read it as a single signed number, not as a minus sign floating next to a 4.
\[ n = -4 \]
Bring the exponent down in front
Why: The coefficient of the answer is the old exponent, sign included.
\[ f'(x) = -4 \cdot x^{\,?} \]
Subtract one from the exponent
Why: Subtracting one from negative four moves further from zero, to negative five. Going down the number line, not up.
\[ -4 - 1 = -5 \quad\Longrightarrow\quad f'(x) = -4x^{-5} = -\frac{4}{x^{5}} \]
Verify numerically at the point where x equals 2
Why: The formula predicts a slope of negative 4 divided by 32, which is negative 0.125. Secant slopes of the original function agree.
| h | secant slope at x = 2 |
|---|---|
| 0.01 | -0.1235 |
| 0.001 | -0.1248 |
| 0.0001 | -0.1250 |
Picture it
Animation
Shows: Each line of the worked example "a negative exponent", 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 negative 4 divided by 32, which is negative 0.125. Secant slopes of the original function agree.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Bring down the 3, then make the exponent negative 2
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The pull is to think 'the exponent gets smaller in size', so negative 3 becomes negative 2.
Same function, and the rule says subtract, not shrink.
Why: The pull is to think 'the exponent gets smaller in size', so negative 3 becomes negative 2.
Trap
Differentiate this function.
\[ f(x) = x^{-3} \]
Bring down the 3, then make the exponent negative 2
Why: The pull is to think 'the exponent gets smaller in size', so negative 3 becomes negative 2.
\[ f'(x) = -3x^{-2} = -\frac{3}{x^{2}} \quad \textbf{(wrong)} \]
Test it at the point where x equals 2
Why: This formula claims a slope of negative 3 over 4, which is negative 0.75. That is four times too steep.
\[ -\frac{3}{2^{2}} = -0.75 \]
Same function, and the rule says subtract, not shrink.
\[ f(x) = x^{-3} \]
Compute the new exponent as negative 3 minus 1
Why: Subtraction is subtraction. Negative three minus one is negative four, which is further from zero than where you started.
\[ f'(x) = -3x^{-4} = -\frac{3}{x^{4}} \]
Test the same point
Why: This predicts negative 3 over 16, which is negative 0.1875.
\[ -\frac{3}{2^{4}} = -0.1875 \]
Check which one the real slopes match
Why: Actual secant slopes of the function at x equal to 2 close in on negative 0.1875, not negative 0.75.
| h | secant slope at x = 2 |
|---|---|
| 0.001 | -0.18731 |
| 0.0001 | -0.18748 |
Trade off
Comparison matrix
From Trap: subtracting one from a negative exponent: every row here is a choice with a cost. Fill the secant slope at x = 2 column, then say which row you would actually pick and what you give up for it.
| h | secant slope at x = 2 |
|---|---|
| 0.001 | -0.18731 |
| 0.0001 | -0.18748 |
Pattern
Predict first
The table runs: 0.1 | 0.16621 · 0.01 | 0.16662
In Worked example: a fractional exponent, given the rows so far: what is the next one — the row where h is 0.001?
Correct: 0.001 | 0.16666
| h | secant slope at x = 9 |
|---|---|
| 0.1 | 0.16621 |
| 0.01 | 0.16662 |
| 0.001 | 0.16666 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The exponent becomes the coefficient, exactly as with a whole number.
Worked example
A square root is a power in disguise. Rewrite it, then the rule applies.
\[ f(x) = \sqrt{x} = x^{1/2} \]
Bring the one half down in front
Why: The exponent becomes the coefficient, exactly as with a whole number.
\[ f'(x) = \tfrac{1}{2}\,x^{\,?} \]
Subtract one from one half
Why: One half minus one is negative one half. Common denominators: two halves make a whole, so one half minus two halves is negative one half.
\[ \tfrac{1}{2} - 1 = -\tfrac{1}{2} \quad\Longrightarrow\quad f'(x) = \tfrac{1}{2}x^{-1/2} \]
Rewrite the answer in root form
Why: A negative exponent means the power belongs in the denominator, and the one half turns back into a square root.
\[ f'(x) = \frac{1}{2\sqrt{x}} \]
Verify numerically at the point where x equals 9
Why: The formula predicts one divided by six, about 0.16667. Secant slopes of the square-root function agree.
| h | secant slope at x = 9 |
|---|---|
| 0.1 | 0.16621 |
| 0.01 | 0.16662 |
| 0.001 | 0.16666 |
Picture it
Animation
Shows: Each line of the worked example "a fractional exponent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula predicts one divided by six, about 0.16667. Secant slopes of the square-root function agree.
Ranking
Put in order
These are the steps of The power-rule recipe, 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
Every power-rule problem is these four moves, in this order.
Then, optionally, clean up: negative exponents can go back into a denominator, fractional ones back into roots.
\[ \frac{d}{dx}\left[c\,x^{n}\right] = c\,n\,x^{\,n-1} \]
Elimination
Eliminate the wrong options
What is the derivative of the function shown?
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: Rewrite as 5 times x to the power negative 3. The power rule gives 5 times negative 3 times x to the power negative 4, which is negative 15 times x to the power negative 4, or negative 15 over x to the 4th.
Check
Rewrite first, then differentiate. Work it on paper before you choose.
\[ f(x) = \frac{5}{x^{3}} \]
Check your understanding
What is the derivative of the function shown?
Answer: A
Why: Rewrite as 5 times x to the power negative 3. The power rule gives 5 times negative 3 times x to the power negative 4, which is negative 15 times x to the power negative 4, or negative 15 over x to the 4th.
Section
Section 2
Concept
A number multiplying a function just rides along through the derivative.
\[ \frac{d}{dx}\left[c \cdot f(x)\right] = c \cdot f'(x) \]
You do not differentiate the constant and you do not drop it. You copy it down and differentiate what it multiplies.
\[ \frac{d}{dx}\left[5x^{4}\right] = 5 \cdot 4x^{3} = 20x^{3} \]
This follows straight from the limit laws: a constant factor can be pulled out of any limit, and the difference quotient is a limit.
Intuition
Multiplying a function by 5 stretches its graph vertically by a factor of 5.
Every rise gets five times bigger. The runs do not change at all. So every slope gets five times bigger too.
| over a run of 1 | original rise | after multiplying by 5 |
|---|---|---|
| from x = 1 to x = 2 | 3 | 15 |
| from x = 2 to x = 3 | 7 | 35 |
Slope is rise over run, so a stretch of five in the rise is a stretch of five in the slope. That is the rule, drawn instead of proved.
Picture it
Animation
Shows: Stretching a graph stretches its slopes — a rendered Manim animation.
Rendered with Manim.
Takeaway: Multiply the function by a constant and every slope scales the same way.
Concept
Derivatives break across addition and subtraction, term by term.
\[ \frac{d}{dx}\left[f(x) \pm g(x)\right] = f'(x) \pm g'(x) \]
This is the rule that turns a long polynomial into a list of small, independent power-rule problems.
It is worth naming what this rule does not say. It says nothing about products and nothing about quotients. Those need their own rules, in the next deck.
Picture it
Animation
Shows: The sum rule marked correct beside the false product shortcut.
The second line is the error to avoid.
Takeaway: Derivatives distribute over sums because rates add. They do not distribute over products, and assuming they do is the commonest mistake here.
Intuition
Suppose water is running into a tank from two hoses. One adds 3 gallons per minute, the other adds 2 gallons per minute.
The total in the tank is the sum of the two contributions, and the total rate is 5 gallons per minute. Nobody is surprised by that.
The sum rule is exactly that statement: if a quantity is a sum of pieces, its rate of change is the sum of the pieces' rates.
Step zero
Discussion prompt
Worked example: a whole polynomial in one pass — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate the first term
Answer:
Worked example
Combine all three rules. Handle each term on its own.
\[ g(x) = 3x^{5} - 7x^{3} + 2x - 9 \]
Differentiate the first term
Why: Constant multiple keeps the 3; the power rule turns x to the 5th into 5 times x to the 4th.
\[ \frac{d}{dx}\left[3x^{5}\right] = 3 \cdot 5x^{4} = 15x^{4} \]
Differentiate the second term, sign and all
Why: The minus belongs to the term. Keep it attached and the difference rule takes care of itself.
\[ \frac{d}{dx}\left[-7x^{3}\right] = -7 \cdot 3x^{2} = -21x^{2} \]
Differentiate the linear term
Why: The derivative of x is 1, so 2 times x has derivative 2. Linear terms leave behind their coefficient.
\[ \frac{d}{dx}\left[2x\right] = 2 \]
Differentiate the constant
Why: The constant rule: it contributes nothing to the slope, so it simply disappears.
\[ \frac{d}{dx}\left[-9\right] = 0 \]
Assemble the pieces
Why: The sum and difference rules let the four separate answers be added back together in the same order.
\[ g'(x) = 15x^{4} - 21x^{2} + 2 \]
Verify numerically at the point where x equals 1
Why: The formula gives 15 minus 21 plus 2, which is negative 4. Secant slopes of the original polynomial close in on negative 4.
| h | secant slope at x = 1 |
|---|---|
| 0.01 | -3.9077 |
| 0.001 | -3.9910 |
| 0.0001 | -3.9991 |
Picture it
Animation
Shows: Each line of the worked example "a whole polynomial in one pass", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives 15 minus 21 plus 2, which is negative 4. Secant slopes of the original polynomial close in on negative 4.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The sum rule worked term by term, so surely a product works factor by factor.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This is the single most tempting false move in first-semester calculus, because the sum rule really does behave this way.
Until you have the product rule, expand first, then use the sum rule.
Why: This is the single most tempting false move in first-semester calculus, because the sum rule really does behave this way.
Trap
The sum rule worked term by term, so surely a product works factor by factor.
\[ f(x) = (2x + 3)(x^{2} - 1) \]
Differentiate each factor and multiply the results
Why: This is the single most tempting false move in first-semester calculus, because the sum rule really does behave this way.
\[ f'(x) \stackrel{?}{=} (2)(2x) = 4x \quad \textbf{(wrong)} \]
Evaluate the claim at the point where x equals 1
Why: This formula predicts a slope of 4 there.
\[ 4(1) = 4 \]
Until you have the product rule, expand first, then use the sum rule.
\[ f(x) = (2x+3)(x^{2}-1) = 2x^{3} + 3x^{2} - 2x - 3 \]
Differentiate the expanded polynomial term by term
Why: Now every term is a constant times a power, which is territory the rules already cover.
\[ f'(x) = 6x^{2} + 6x - 2 \]
Evaluate at the same point
Why: This predicts 6 plus 6 minus 2, which is 10. The two answers are not close.
\[ 6(1)^2 + 6(1) - 2 = 10 \]
Check which prediction the real slopes match
Why: Actual secant slopes of the product at x equal to 1 head for 10, not 4. Differentiating factor by factor is simply false.
| h | secant slope at x = 1 |
|---|---|
| 0.01 | 10.0902 |
| 0.001 | 10.0090 |
| 0.0001 | 10.0009 |
Pattern
Step through it
Step through Trap: the derivative does not distribute over a product one row at a time. What is driving the change, and what would the row after the last one be?
Fill the middle
Fill in the blanks
From Worked example: expand, then differentiate — finish the line. Write what belongs on the right of the equals sign before you look.
f(x) = 3x^{3} - 4x^{2}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Multiplying powers of the same base adds exponents, so x squared times 3x is 3 times x cubed.
Worked example
When a product is small enough to multiply out, multiplying out is the fastest legal route.
\[ f(x) = x^{2}\left(3x - 4\right) \]
Distribute the outside factor
Why: Multiplying powers of the same base adds exponents, so x squared times 3x is 3 times x cubed.
\[ f(x) = 3x^{3} - 4x^{2} \]
Differentiate each term
Why: Constant multiple plus power rule on each piece, then the difference rule to join them.
\[ f'(x) = 9x^{2} - 8x \]
Check the tempting wrong answer for contrast
Why: Factor-by-factor would have given 2x times 3, which is 6x. At x equal to 2 that says 12, while the real answer says 36 minus 16, which is 20.
\[ f'(2) = 9(4) - 8(2) = 20 \]
Verify numerically at the point where x equals 2
Why: Secant slopes of the original product form close in on 20, confirming the expanded route.
| h | secant slope at x = 2 |
|---|---|
| 0.01 | 20.1403 |
| 0.001 | 20.0140 |
| 0.0001 | 20.0014 |
Picture it
Animation
Shows: Each line of the worked example "expand, then differentiate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Factor-by-factor would have given 2x times 3, which is 6x. At x equal to 2 that says 12, while the real answer says 36 minus 16, which is 20.
Pattern
Any polynomial, every time:
Two warnings that pay for themselves: a product is not a sum, and a quotient is not a sum. Expand or rewrite before you split.
Prediction
Predict first
What is the derivative of the polynomial shown?
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: 12x squared minus 12x plus 7
Why: Term by term: 4 times 3 gives 12x squared, negative 6 times 2 gives negative 12x, the linear term 7x leaves behind its coefficient 7, and the constant negative 12 differentiates to zero.
Check
Term by term. Watch the constant and watch the linear term.
\[ f(x) = 4x^{3} - 6x^{2} + 7x - 12 \]
Check your understanding
What is the derivative of the polynomial shown?
Answer: A
Why: Term by term: 4 times 3 gives 12x squared, negative 6 times 2 gives negative 12x, the linear term 7x leaves behind its coefficient 7, and the constant negative 12 differentiates to zero.
Section
Section 3
Concept
The power rule needs to see an exponent. A radical sign hides one.
\[ \sqrt[n]{x^{\,m}} = x^{\,m/n} \]
| written as a root | written as a power |
|---|---|
| square root of x | x to the one half |
| cube root of x | x to the one third |
| cube root of x squared | x to the two thirds |
| fourth root of x cubed | x to the three fourths |
Reading the table left to right is the setup move. Reading it right to left is how you clean up the answer at the end.
Intuition
Think of the power rule as a machine with one slot, labelled 'exponent'. It cannot read a radical sign, and it cannot read a fraction bar.
Rewriting is not extra work you do to be tidy. It is the step that gets the expression into the machine at all.
Almost every 'I do not know which rule to use' moment in this chapter is really a 'I have not rewritten it yet' moment.
Hypothesis
Predict first
Worked example: a cube root 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: Rewrite the root as a power
Why: A cube root is the one-third power. The coefficient 4 rides along untouched by the constant multiple rule.
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
Rewrite, differentiate, then translate the answer back into root form.
\[ f(x) = 4\sqrt[3]{x} \]
Rewrite the root as a power
Why: A cube root is the one-third power. The coefficient 4 rides along untouched by the constant multiple rule.
\[ f(x) = 4x^{1/3} \]
Apply the power rule to the variable part
Why: Bring one third down in front, then subtract one from the exponent: one third minus three thirds is negative two thirds.
\[ f'(x) = 4 \cdot \tfrac{1}{3}\,x^{-2/3} = \tfrac{4}{3}x^{-2/3} \]
Clean up into root form
Why: The negative exponent sends the power to the denominator, and two thirds is the cube root of the square.
\[ f'(x) = \frac{4}{3\sqrt[3]{x^{2}}} \]
Evaluate at a friendly point
Why: At x equal to 8, the cube root of 8 squared is 4, so the slope should be 4 divided by 12, which is one third.
\[ f'(8) = \frac{4}{3 \cdot 4} = \frac{1}{3} \]
Verify numerically at the point where x equals 8
Why: Secant slopes of the original cube-root function close in on 0.33333, which is one third.
| h | secant slope at x = 8 |
|---|---|
| 0.1 | 0.33195 |
| 0.01 | 0.33319 |
| 0.001 | 0.33332 |
Picture it
Animation
Shows: Each line of the worked example "a cube root", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Secant slopes of the original cube-root function close in on 0.33333, which is one third.
Concept
A variable sitting under a fraction bar comes upstairs with the sign of its exponent flipped.
\[ \frac{1}{x^{\,n}} = x^{-n} \]
| written as a fraction | written as a power |
|---|---|
| 1 over x | x to the power negative 1 |
| 3 over x squared | 3 times x to the power negative 2 |
| 1 over the square root of x | x to the power negative one half |
| 1 over the quantity 2x | one half times x to the power negative 1 |
Watch the last row. Only the variable moves. A constant factor in the denominator becomes a fraction out front and stays there.
Comparison
Comparison matrix
From Rewrite move two: a denominator is a negative power: refill the written as a power column from what you know. The rest of the table is as it appeared.
| written as a fraction | written as a power |
|---|---|
| 1 over x | x to the power negative 1 |
| 3 over x squared | 3 times x to the power negative 2 |
| 1 over the square root of x | x to the power negative one half |
| 1 over the quantity 2x | one half times x to the power negative 1 |
Intuition
A negative exponent is not a negative number and it does not make anything negative. It is a location marker: it says 'this power belongs downstairs'.
Moving a power across the fraction bar in either direction flips the sign of its exponent. That is the entire content of the notation.
So there is never a reason to be stuck on a reciprocal. Bring it upstairs, differentiate, and send whatever is left back down.
Ranking
Put in order
Put the moves of Worked example: two reciprocal terms 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 3 stays as a coefficient; the 2 in the second denominator becomes the coefficient one half.
Worked example
Neither term is ready for the power rule as written. Both are one rewrite away.
\[ f(x) = \frac{3}{x^{2}} + \frac{1}{2x} \]
Rewrite both terms as powers
Why: The 3 stays as a coefficient; the 2 in the second denominator becomes the coefficient one half. Only the variable changes floors.
\[ f(x) = 3x^{-2} + \tfrac{1}{2}x^{-1} \]
Differentiate the first term
Why: Bring negative 2 down and multiply it by the 3; the new exponent is negative 2 minus 1, which is negative 3.
\[ \frac{d}{dx}\left[3x^{-2}\right] = -6x^{-3} \]
Differentiate the second term
Why: Bring negative 1 down and multiply it by one half; the new exponent is negative 1 minus 1, which is negative 2.
\[ \frac{d}{dx}\left[\tfrac{1}{2}x^{-1}\right] = -\tfrac{1}{2}x^{-2} \]
Add the pieces and send the powers back downstairs
Why: The sum rule joins them; the negative exponents mean both terms belong under fraction bars.
\[ f'(x) = -\frac{6}{x^{3}} - \frac{1}{2x^{2}} \]
Verify numerically at the point where x equals 1
Why: The formula gives negative 6 minus one half, which is negative 6.5. Secant slopes of the original expression close in on negative 6.5.
| h | secant slope at x = 1 |
|---|---|
| 0.01 | -6.40623 |
| 0.001 | -6.49051 |
| 0.0001 | -6.49905 |
Picture it
Animation
Shows: Each line of the worked example "two reciprocal terms", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives negative 6 minus one half, which is negative 6.5. Secant slopes of the original expression close in on negative 6.5.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The expression looks like it has no exponents to work with, so people improvise.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The improvised move: turn x to the 4th into 4 times x cubed downstairs, and decide the cube root has 'no exponent' so it contributes nothing.
Rewrite both terms as powers first. Then there is nothing to improvise.
Why: The improvised move: turn x to the 4th into 4 times x cubed downstairs, and decide the cube root has 'no exponent' so it contributes nothing.
Trap
The expression looks like it has no exponents to work with, so people improvise.
\[ f(x) = \frac{3}{x^{4}} + \sqrt[3]{x} \]
Differentiate the denominator of the first term and leave the root alone
Why: The improvised move: turn x to the 4th into 4 times x cubed downstairs, and decide the cube root has 'no exponent' so it contributes nothing.
\[ f'(x) \stackrel{?}{=} \frac{3}{4x^{3}} \quad \textbf{(wrong)} \]
Test it at the point where x equals 1
Why: This claims the slope there is 3 divided by 4, a small positive number.
\[ \frac{3}{4(1)^{3}} = 0.75 \]
Rewrite both terms as powers first. Then there is nothing to improvise.
\[ f(x) = 3x^{-4} + x^{1/3} \]
Power rule on each term
Why: First term: negative 4 times 3 gives negative 12, exponent negative 5. Second term: one third out front, exponent one third minus one, which is negative two thirds.
\[ f'(x) = -12x^{-5} + \tfrac{1}{3}x^{-2/3} = -\frac{12}{x^{5}} + \frac{1}{3\sqrt[3]{x^{2}}} \]
Test the same point
Why: At x equal to 1 this gives negative 12 plus one third, about negative 11.667. The sign alone is opposite to the improvised answer.
\[ f'(1) = -12 + \tfrac{1}{3} \approx -11.667 \]
Check which prediction the real slopes match
Why: Actual secant slopes are large and negative, heading for about negative 11.667. The function is falling steeply there, not rising gently.
| h | secant slope at x = 1 |
|---|---|
| 0.001 | -11.6368 |
| 0.0001 | -11.6637 |
| 0.00001 | -11.6664 |
Pattern
Step through it
Step through Trap: differentiating before rewriting one row at a time. What is driving the change, and what would the row after the last one be?
Concept
When several terms share a single denominator, hand the denominator to each term separately.
\[ \frac{a + b + c}{d} = \frac{a}{d} + \frac{b}{d} + \frac{c}{d} \]
This is legal only because the denominator is a single term. Splitting across a sum in the denominator is not a thing and never has been.
\[ \frac{1}{a+b} \neq \frac{1}{a} + \frac{1}{b} \]
After the split, each piece simplifies to a power, and the sum rule takes over.
Picture it
Animation
Shows: Split a sum over one denominator — a rendered Manim animation.
Rendered with Manim.
Takeaway: No quotient rule required, and no chance of a sign error.
Estimation
Predict first
One fraction bar, three terms on top, a single term on the bottom. Split it.
Commit before you compute: what does Worked example: split, then differentiate 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 point where x equals 1
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. Secant slopes of the original single-fraction form close in on negative 1, so the split did not change the function.
Worked example
One fraction bar, three terms on top, a single term on the bottom. Split it.
\[ f(x) = \frac{2x^{4} - x^{2} + 5x}{x^{2}} \]
Give the denominator to each term
Why: The denominator is one term, so this split is legal. Now each piece is a quotient of powers.
\[ f(x) = \frac{2x^{4}}{x^{2}} - \frac{x^{2}}{x^{2}} + \frac{5x}{x^{2}} \]
Simplify each piece by subtracting exponents
Why: Dividing powers of the same base subtracts exponents: 4 minus 2 is 2, 2 minus 2 is 0, and 1 minus 2 is negative 1.
\[ f(x) = 2x^{2} - 1 + 5x^{-1} \]
Differentiate term by term
Why: Power rule on the first, constant rule on the middle, power rule with a negative exponent on the last.
\[ f'(x) = 4x - 5x^{-2} = 4x - \frac{5}{x^{2}} \]
Evaluate at the point where x equals 1
Why: The formula gives 4 minus 5, which is negative 1. A negative slope on a function whose value there is 6.
\[ f'(1) = 4 - 5 = -1 \]
Verify numerically at the point where x equals 1
Why: Secant slopes of the original single-fraction form close in on negative 1, so the split did not change the function.
| h | secant slope at x = 1 |
|---|---|
| 0.01 | -0.93050 |
| 0.001 | -0.99301 |
| 0.0001 | -0.99930 |
Picture it
Animation
Shows: Each line of the worked example "split, then differentiate", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Secant slopes of the original single-fraction form close in on negative 1, so the split did not change the function.
Constraint
Discussion prompt
Run The rewrite-first checklist with this step confiscated:
A sum over a single denominator? Split it and simplify each piece.
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
Before you differentiate anything, run this list. It takes ten seconds and saves the problem.
The goal of every rewrite is the same shape: a sum of terms, each a number times a power of the variable. Get there and the problem is finished.
Edge cases
Discussion prompt
The rewrite-first checklist 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:
Before you differentiate anything, run this list. It takes ten seconds and saves the problem.
Check
Two rewrites are needed here, one per term.
\[ f(x) = \sqrt{x} + \frac{2}{x} \]
Check your understanding
What is the derivative of the function shown?
Answer: A
Why: Rewrite as x to the one half plus 2 times x to the power negative 1. The power rule gives one half times x to the power negative one half, plus 2 times negative 1 times x to the power negative 2, which is 1 over 2 root x minus 2 over x squared.
Section
Section 4
Concept
One more rule to add to the toolkit, and it is the strangest one in calculus.
\[ \frac{d}{dx}\left[e^{x}\right] = e^{x} \]
natural exponential function — The exponential function whose base is the number e, about 2.71828. It is the unique exponential function that is exactly equal to its own derivative at every point.
Take it as given here. The proof needs a limit you will meet again with logarithms, and it is not what makes the rule useful.
Definition probe
Sort into buckets
Every line below is part of the definition of constant rule or of natural exponential function — one or the other, never both. Put each where it belongs.
Intuition
Picture the graph. At the height 1, the graph is climbing at a rate of 1 unit per unit. At the height 7.389, it is climbing at a rate of 7.389.
| x | value of the function | slope there |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 2.71828 | 2.71828 |
| 2 | 7.38906 | 7.38906 |
The taller it gets, the faster it grows, in exactly the same proportion. That is what runaway growth means, and it is why this function shows up in population, interest, and decay problems.
The number e is not chosen for elegance. It is the one base that makes the slope match the height exactly, with no fudge factor.
Pattern
Step through it
Step through The function that is its own slope one row at a time. What is driving the change, and what would the row after the last one be?
Anomaly
Predict first
A student writes this, and it looks reasonable:
The expression has a base and an exponent, so the power rule feels like it should apply.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This copies the power rule blindly.
Check which slot the variable is in. Here the variable is the exponent, so this is an exponential function, not a power function.
Why: This copies the power rule blindly. But the power rule was proved for a variable base with a constant exponent, which is the opposite arrangement.
Trap
The expression has a base and an exponent, so the power rule feels like it should apply.
\[ f(x) = 2^{x} \]
Bring the exponent down and subtract one from it
Why: This copies the power rule blindly. But the power rule was proved for a variable base with a constant exponent, which is the opposite arrangement.
\[ f'(x) \stackrel{?}{=} x \cdot 2^{\,x-1} \quad \textbf{(wrong)} \]
Test it at the point where x equals 3
Why: This claims a slope of 3 times 2 squared, which is 12.
\[ 3 \cdot 2^{2} = 12 \]
Check which slot the variable is in. Here the variable is the exponent, so this is an exponential function, not a power function.
\[ f(x) = 2^{x} \]
Use the exponential rule for a general base
Why: Every exponential can be written with base e, and doing so leaves behind a factor of the natural log of the base. The full derivation comes in the transcendental-derivatives deck.
\[ \frac{d}{dx}\left[2^{x}\right] = 2^{x}\ln 2 \]
Test the same point
Why: At x equal to 3 this gives 8 times the natural log of 2, about 5.5452 - less than half of what the wrong rule claimed.
\[ 2^{3}\ln 2 = 8(0.69315) \approx 5.5452 \]
Check which prediction the real slopes match
Why: Actual secant slopes head for about 5.5452, nowhere near 12. The test to remember: variable in the base means power rule, variable in the exponent means exponential rule.
| h | secant slope at x = 3 |
|---|---|
| 0.1 | 5.7418 |
| 0.01 | 5.5644 |
| 0.001 | 5.5471 |
Pattern
Step through it
Step through Trap: the power rule on a constant base one row at a time. What is driving the change, and what would the row after the last one be?
Step zero
Discussion prompt
Worked example: powers and the exponential together — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate the exponential term
Answer:
Worked example
Nothing new. The sum rule lets the exponential term sit beside the power terms.
\[ f(x) = 3e^{x} + x^{3} - 2x \]
Differentiate the exponential term
Why: The constant multiple rule keeps the 3, and the exponential is unchanged by differentiation.
\[ \frac{d}{dx}\left[3e^{x}\right] = 3e^{x} \]
Differentiate the power term and the linear term
Why: Power rule on the cube gives 3 times x squared; the linear term leaves behind its coefficient negative 2.
\[ \frac{d}{dx}\left[x^{3} - 2x\right] = 3x^{2} - 2 \]
Assemble
Why: The sum and difference rules join the three answers in place.
\[ f'(x) = 3e^{x} + 3x^{2} - 2 \]
Evaluate at the point where x equals 0
Why: The exponential is 1 there, so the slope is 3 times 1, plus 0, minus 2.
\[ f'(0) = 3(1) + 0 - 2 = 1 \]
Verify numerically at the point where x equals 0
Why: Secant slopes of the original function close in on 1, confirming that the exponential term contributed exactly 3.
| h | secant slope at x = 0 |
|---|---|
| 0.1 | 1.16513 |
| 0.01 | 1.01515 |
| 0.001 | 1.00150 |
Picture it
Animation
Shows: Each line of the worked example "powers and the exponential together", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Secant slopes of the original function close in on 1, confirming that the exponential term contributed exactly 3.
Commit first
Predict first
What is the derivative of the function shown?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: 2x plus 2 to the x times the natural log of 2
Why: The first term has the variable in the base, so the power rule gives 2x. The second term has the variable in the exponent, so the exponential rule gives 2 to the x times the natural log of 2, about 0.69315 times 2 to the 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 term is a power function and one is an exponential function. Sort them before you differentiate.
\[ f(x) = x^{2} + 2^{x} \]
Check your understanding
What is the derivative of the function shown?
Answer: A
Why: The first term has the variable in the base, so the power rule gives 2x. The second term has the variable in the exponent, so the exponential rule gives 2 to the x times the natural log of 2, about 0.69315 times 2 to the x.
Concept
The derivative of a function is itself a function, so it can be differentiated again.
\[ f''(x) = \frac{d}{dx}\left[f'(x)\right] \]
second derivative — The derivative of the derivative. It measures how fast the slope itself is changing, which is why it controls the bend of a graph.
You can keep going: third, fourth, and beyond. For a polynomial the process always terminates at zero, and that is a useful sanity check.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of constant rule, natural exponential function, second derivative as Power, Constant, Sum, and Difference Rules uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Three dashboard readings, and each one is the rate of change of the one before it.
| quantity | what it reads | derivative of |
|---|---|---|
| position | where the car is | - |
| velocity | how fast the position is changing | position |
| acceleration | how fast the velocity is changing | velocity |
Pressing the gas does not move the car directly. It changes the speed, and the changing speed moves the car. That is a second derivative acting through a first one.
This is also why acceleration can be negative while the car is still moving forward: the speed is dropping, but it has not reached zero yet.
Picture it
Animation
Shows: Odometer, speedometer, gas pedal — a rendered Manim animation.
Rendered with Manim.
Takeaway: Each derivative answers a question about the one before it.
Concept
Three notations, all standard, all meaning the same thing.
\[ f''(x) \qquad \frac{d^{2}y}{dx^{2}} \qquad y'' \]
After the third derivative, primes get unreadable, so a bracketed number takes over.
\[ f^{(4)}(x) \qquad \frac{d^{4}y}{dx^{4}} \]
The parentheses matter. A bracketed 4 up there means the fourth derivative; a bare 4 would mean the fourth power. They are completely different objects.
Concept
The first derivative answers 'is it going up or down?'. The second derivative answers 'is that changing?'.
| first derivative | second derivative | what the graph is doing |
|---|---|---|
| positive | positive | rising, and getting steeper |
| positive | negative | rising, but flattening out |
| negative | negative | falling, and getting steeper |
| negative | positive | falling, but levelling off |
You will use this table constantly in the curve-sketching deck. For now, notice that the sign of the second derivative is about bend, not about direction.
Trade off
Comparison matrix
From What the second derivative tells you: every row here is a choice with a cost. Fill the what the graph is doing column, then say which row you would actually pick and what you give up for it.
| first derivative | second derivative | what the graph is doing |
|---|---|---|
| positive | positive | rising, and getting steeper |
| positive | negative | rising, but flattening out |
| negative | negative | falling, and getting steeper |
| negative | positive | falling, but levelling off |
Picture it
Animation
Shows: What the second derivative tells you — a rendered Manim animation.
Rendered with Manim.
Takeaway: Where the bend switches direction, the second derivative changes sign.
Pattern
Predict first
The table runs: 0.1 | -5.0200 · 0.01 | -5.9092
In Worked example: every derivative of a quartic, given the rows so far: what is the next one — the row where h is 0.001?
Correct: 0.001 | -5.9910
| h | secant slope of the first derivative at x = 1 |
|---|---|
| 0.1 | -5.0200 |
| 0.01 | -5.9092 |
| 0.001 | -5.9910 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Power rule on each term; the constant negative 7 disappears.
Worked example
Differentiate repeatedly and watch the degree drop by one each time.
\[ f(x) = 2x^{4} - 5x^{3} + x - 7 \]
First derivative
Why: Power rule on each term; the constant negative 7 disappears.
\[ f'(x) = 8x^{3} - 15x^{2} + 1 \]
Second derivative
Why: Differentiate the first derivative. The constant 1 that survived the first round now disappears in its turn.
\[ f''(x) = 24x^{2} - 30x \]
Third and fourth derivatives
Why: Each pass drops the degree by one, so the fourth derivative of a fourth-degree polynomial is a constant.
\[ f'''(x) = 48x - 30 \qquad f^{(4)}(x) = 48 \]
Fifth derivative and beyond
Why: The fourth derivative is constant, so everything after it is zero forever. Degree four, four steps to a constant: that is the sanity check.
\[ f^{(5)}(x) = 0 \]
Verify the second derivative at the point where x equals 1
Why: The formula gives 24 minus 30, which is negative 6. Taking secant slopes of the FIRST derivative near x equal to 1 must reproduce that number, and it does.
| h | secant slope of the first derivative at x = 1 |
|---|---|
| 0.1 | -5.0200 |
| 0.01 | -5.9092 |
| 0.001 | -5.9910 |
Picture it
Animation
Shows: Each line of the worked example "every derivative of a quartic", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives 24 minus 30, which is negative 6. Taking secant slopes of the FIRST derivative near x equal to 1 must reproduce that number, and it does.
Missing information
Discussion prompt
A particle moves along a line. Position is in meters and time is in seconds.
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:
Velocity is the rate of change of position, so its units are meters per second.
Worked example
A particle moves along a line. Position is in meters and time is in seconds.
\[ s(t) = t^{3} - 6t^{2} + 9t \]
Differentiate once for velocity
Why: Velocity is the rate of change of position, so its units are meters per second.
\[ v(t) = s'(t) = 3t^{2} - 12t + 9 \]
Differentiate again for acceleration
Why: Acceleration is the rate of change of velocity, so its units are meters per second per second.
\[ a(t) = v'(t) = 6t - 12 \]
Find when the particle is momentarily at rest
Why: At rest means velocity zero. Factor out the 3 and the quadratic factors cleanly.
\[ 3t^{2} - 12t + 9 = 3(t-1)(t-3) = 0 \;\Longrightarrow\; t = 1,\; t = 3 \]
Read the motion off the two functions
Why: Velocity is positive before 1 second, negative between 1 and 3, positive again after 3. Acceleration is zero at 2 seconds, which is exactly where the velocity bottoms out.
| t (s) | position (m) | velocity (m/s) | acceleration (m/s per s) |
|---|---|---|---|
| 0 | 0 | 9 | -12 |
| 1 | 4 | 0 | -6 |
| 2 | 2 | -3 | 0 |
| 3 | 0 | 0 | 6 |
| 4 | 4 | 9 | 12 |
Verify the turning points against the position values
Why: Position rises to 4 at 1 second, falls to 0 at 3 seconds, then rises again - exactly the two places the velocity crossed zero. The two derivatives tell a story the position table confirms.
Picture it
Animation
Shows: Each line of the worked example "position, velocity, acceleration", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Position rises to 4 at 1 second, falls to 0 at 3 seconds, then rises again - exactly the two places the velocity crossed zero. The two derivatives tell a story the position table confirms.
Check
Differentiate twice, then substitute. Do not substitute early.
\[ f(x) = x^{4} - 2x^{3} + 5x, \qquad \text{find } f''(2) \]
Check your understanding
What is the value of the second derivative at the point where x equals 2?
Answer: A
Why: The first derivative is 4x cubed minus 6x squared plus 5. The second derivative is 12x squared minus 12x. Substituting 2 gives 48 minus 24, which is 24.
Section
Section 5
Picture it
Figure (svg): A curve with a straight dashed line touching it at a single marked point.
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:
Every derivative you computed in this deck was a slope formula. Feed it a number and it hands back the steepness of the curve there.
Concept
Every derivative you computed in this deck was a slope formula. Feed it a number and it hands back the steepness of the curve there.
\[ m = f'(a) \]
Figure (svg): A curve with a straight dashed line touching it at a single marked point.
The tangent line at a point is the straight line that agrees with the curve in two ways at once: it passes through the same point, and it has the same slope.
Intuition
To write down any line you need a point on it and its slope. That is it. Point-slope form is built for exactly this.
\[ y - y_{1} = m\,(x - x_{1}) \]
The function supplies the point. The derivative supplies the slope. Two different formulas, evaluated at the same input.
That split is where most tangent-line mistakes come from: people evaluate one formula twice instead of two formulas once each.
Explain it
Discussion prompt
Explain A line needs exactly two ingredients 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:
To write down any line you need a point on it and its slope. That is it. Point-slope form is built for exactly this.
Picture it
Animation
Shows: A tangent line settling onto a cubic at a chosen point.
Slope from the derivative, point from the function.
Takeaway: The derivative supplies the slope and the original function supplies the point. Those two ingredients determine the tangent line completely.
Estimation
Predict first
Find the equation of the line tangent to this curve at the point where the input is 2.
Commit before you compute: what does Worked example: find the tangent line 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 that the line touches the curve and matches its slope
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. At x equal to 2 the line gives 16 minus 15, which is 1 - the same as the function.
Worked example
Find the equation of the line tangent to this curve at the point where the input is 2.
\[ f(x) = x^{3} - 4x + 1 \]
Get the point from the original function
Why: The tangent line has to touch the curve, so it must pass through this point. Use the function, not the derivative.
\[ f(2) = 8 - 8 + 1 = 1 \;\Longrightarrow\; (2,\,1) \]
Differentiate
Why: Power rule on the cube, the linear term leaves its coefficient, the constant vanishes.
\[ f'(x) = 3x^{2} - 4 \]
Get the slope from the derivative
Why: Now use the second formula at the same input. This number is the slope, not a point on the graph.
\[ f'(2) = 3(4) - 4 = 8 \]
Assemble in point-slope form, then simplify
Why: Substitute the point and the slope, then distribute and solve for y to get slope-intercept form.
\[ y - 1 = 8(x - 2) \;\Longrightarrow\; y = 8x - 15 \]
Verify that the line touches the curve and matches its slope
Why: At x equal to 2 the line gives 16 minus 15, which is 1 - the same as the function. And near that point the line and the curve stay close, which is what tangency looks like numerically.
| x | curve value | tangent line value |
|---|---|---|
| 1.9 | 0.259 | 0.200 |
| 2.0 | 1.000 | 1.000 |
| 2.1 | 1.861 | 1.800 |
Picture it
Animation
Shows: Each line of the worked example "find the tangent line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 2 the line gives 16 minus 15, which is 1 - the same as the function. And near that point the line and the curve stay close, which is what tangency looks like numerically.
Ranking
Put in order
These are the steps of The tangent-line recipe, 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
Four steps, always in this order:
The single most common error is swapping steps one and three: using the derivative's value as the y-coordinate, or the function's value as the slope. Label them as you go.
Prediction
Predict first
What is the equation of the tangent line at the point where x equals 2?
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: y equals negative 0.25x plus 1
Why: The point is given by the function: one half, so the point is (2, 0.5). The derivative of x to the power negative 1 is negative x to the power negative 2, which is negative 0.25 at x equal to 2. Point-slope gives y minus 0.5 equals negative 0.25 times the quantity x minus 2, so y equals negative 0.25x plus 1.
Check
Rewrite first, then run the recipe. Keep the point and the slope in separate boxes.
\[ f(x) = \frac{1}{x}, \qquad \text{tangent at } x = 2 \]
Check your understanding
What is the equation of the tangent line at the point where x equals 2?
Answer: A
Why: The point is given by the function: one half, so the point is (2, 0.5). The derivative of x to the power negative 1 is negative x to the power negative 2, which is negative 0.25 at x equal to 2. Point-slope gives y minus 0.5 equals negative 0.25 times the quantity x minus 2, so y equals negative 0.25x plus 1.
Concept
A horizontal line has slope zero. So asking where a curve has a horizontal tangent is asking where its derivative equals zero.
\[ \text{horizontal tangent at } x = a \iff f'(a) = 0 \]
This converts a question about a picture into an ordinary equation to solve. That trade is one of the most useful moves in the whole course.
These points are where a graph levels off - the tops of hills and the bottoms of valleys. Finding them is the core of the optimization deck later on.
Analogy
Discussion prompt
Explain Horizontal tangents 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:
A horizontal line has slope zero. So asking where a curve has a horizontal tangent is asking where its derivative equals zero.
Picture it
Animation
Shows: Where the curve levels off — a rendered Manim animation.
Rendered with Manim.
Takeaway: Set the derivative to zero and you are asking where the tangent is flat.
Picture it
Figure (svg): A wavy curve with a peak and a valley, each marked with a short dashed horizontal segment and a dot.
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:
Walk along the curve from left to right. You climb, you stop climbing for an instant at the summit, and then you start descending.
Intuition
Figure (svg): A wavy curve with a peak and a valley, each marked with a short dashed horizontal segment and a dot.
Walk along the curve from left to right. You climb, you stop climbing for an instant at the summit, and then you start descending.
That instant of neither climbing nor descending is the horizontal tangent. The slope has to pass through zero to change sign.
Counterexample
Discussion prompt
Walk along the curve from left to right. You climb, you stop climbing for an instant at the summit, and then you start descending.
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.
Step zero
Discussion prompt
Worked example: find every horizontal tangent — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate
Answer:
Worked example
Find all points where the tangent line to this curve is horizontal.
\[ f(x) = x^{3} - 3x^{2} - 9x + 5 \]
Differentiate
Why: Power rule term by term; the constant 5 drops out.
\[ f'(x) = 3x^{2} - 6x - 9 \]
Set the derivative equal to zero
Why: Horizontal tangent means slope zero, and the derivative is the slope formula.
\[ 3x^{2} - 6x - 9 = 0 \]
Factor out the common 3, then factor the quadratic
Why: Pulling out the 3 first makes the trinomial monic and easy to factor: two numbers multiplying to negative 3 and adding to negative 2.
\[ 3\left(x^{2} - 2x - 3\right) = 3(x-3)(x+1) = 0 \]
Solve, then get the y-coordinates from the ORIGINAL function
Why: The derivative gives the x-values. Only the original function can give the heights of those points.
\[ x = 3:\; f(3) = -22 \qquad x = -1:\; f(-1) = 10 \]
Verify by substituting both x-values into the derivative
Why: At x equal to 3: 27 minus 18 minus 9 is 0. At x equal to negative 1: 3 plus 6 minus 9 is 0. Both slopes really are zero, so the points are (3, -22) and (-1, 10).
| x | value of the derivative | point on the curve |
|---|---|---|
| 3 | 0 | (3, -22) |
| -1 | 0 | (-1, 10) |
Picture it
Animation
Shows: Each line of the worked example "find every horizontal tangent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 3: 27 minus 18 minus 9 is 0. At x equal to negative 1: 3 plus 6 minus 9 is 0. Both slopes really are zero, so the points are (3, -22) and (-1, 10).
Ranking
Put in order
Put the moves of Worked example: where does the slope equal a given number? 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 derivative is the slope formula, so it is the thing that gets set equal to 9.
Worked example
Same machinery, different right-hand side. Find every point where the tangent has slope 9.
\[ f(x) = x^{3} - 3x \]
Differentiate
Why: The derivative is the slope formula, so it is the thing that gets set equal to 9.
\[ f'(x) = 3x^{2} - 3 \]
Set the slope formula equal to 9
Why: Asking 'where is the slope 9' is literally asking which inputs make the derivative output 9.
\[ 3x^{2} - 3 = 9 \]
Solve the equation
Why: Add 3, divide by 3, then take both square roots. Forgetting the negative root loses half the answer.
\[ 3x^{2} = 12 \;\Longrightarrow\; x^{2} = 4 \;\Longrightarrow\; x = 2 \text{ or } x = -2 \]
Find the two points
Why: Heights come from the original function: 8 minus 6 is 2, and negative 8 plus 6 is negative 2.
\[ (2,\,2) \quad \text{and} \quad (-2,\,-2) \]
Verify both slopes in the derivative
Why: At x equal to 2: 3 times 4 minus 3 is 9. At x equal to negative 2: 3 times 4 minus 3 is 9 again, because squaring erased the sign. Both points check out.
| x | value of the derivative |
|---|---|
| 2 | 9 |
| -2 | 9 |
Picture it
Animation
Shows: Each line of the worked example "where does the slope equal a given number?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to 2: 3 times 4 minus 3 is 9. At x equal to negative 2: 3 times 4 minus 3 is 9 again, because squaring erased the sign. Both points check out.
Elimination
Eliminate the wrong options
At which x-values does this curve have a horizontal tangent?
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 derivative is 3x squared minus 12. Setting it equal to zero gives x squared equal to 4, so x equals 2 or x equals negative 2. Both values make the slope zero.
Check
Set the derivative to zero and solve. Watch for more than one answer.
\[ f(x) = x^{3} - 12x + 1 \]
Check your understanding
At which x-values does this curve have a horizontal tangent?
Answer: A
Why: The derivative is 3x squared minus 12. Setting it equal to zero gives x squared equal to 4, so x equals 2 or x equals negative 2. Both values make the slope zero.
Pattern
Every rule from this deck, in one place.
| function | derivative | name |
|---|---|---|
| a constant c | 0 | constant rule |
| x to the n | n times x to the n minus 1 | power rule |
| c times f | c times the derivative of f | constant multiple |
| f plus or minus g | derivative of f plus or minus derivative of g | sum and difference |
| e to the x | e to the x | natural exponential |
And the two questions to ask before you start:
Comparison
Comparison matrix
From The whole toolkit on one card: refill the derivative column from what you know. The rest of the table is as it appeared.
| function | derivative | name |
|---|---|---|
| a constant c | 0 | constant rule |
| x to the n | n times x to the n minus 1 | power rule |
| c times f | c times the derivative of f | constant multiple |
| f plus or minus g | derivative of f plus or minus derivative of g | sum and difference |
| e to the x | e to the x | natural exponential |
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — From the Limit to a Rule · Building Polynomials Out of Powers · Rewrite Before You Differentiate · The Exponential and Higher Derivatives · Tangent Lines and Flat Spots. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You have traded the limit definition for a toolkit, and you know when each tool applies.
The four errors to keep watching for:
Next up: the product and quotient rules, which finally handle the products you had to expand by hand in this deck - and then the chain rule, which handles everything else.
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