This deck shows how to differentiate a curve you cannot solve for y, and how to tame ugly products, quotients, and variable exponents by taking a logarithm first. It targets the four classic errors: differentiating a y term as if it were a constant, skipping the product rule on an x-times-y term, failing to collect every dy/dx term before dividing, and forcing the power rule onto a variable base with a variable exponent.
Subject: Calculus I · 142 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 10
What to do when you cannot solve for y, and when the product rule is not enough.
Objectives
By the end of this deck you can:
dy/dx factor to every y term.dy/dx, then evaluate it at a point on the curve.Everything here is one idea used twice: y is secretly a function of x, so the chain rule fires on it.
Warm-up
Discussion prompt
Before we open Implicit and Logarithmic Differentiation: without looking back, what was the main idea of Derivatives of Trig, Exponential, and Log Functions, 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 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.
Section
Part 1
Concept
Every derivative you have taken so far came pre-solved: y sat alone on the left, and a recipe in x sat on the right.
This one does not.
\[ x^2 + y^2 = 25 \]
It is a perfectly good curve - the circle of radius five centered at the origin. It just is not a function of x.
Counterexample
Discussion prompt
Every derivative you have taken so far came pre-solved: y sat alone on the left, and a recipe in x sat on the right.
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
Figure (svg): A circle centered at the origin with a vertical dashed line crossing it at two marked points
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:
A vertical line at almost any x crosses the circle twice. Two outputs for one input means it fails the vertical line test.
Intuition
Figure (svg): A circle centered at the origin with a vertical dashed line crossing it at two marked points
A vertical line at almost any x crosses the circle twice. Two outputs for one input means it fails the vertical line test.
Every differentiation rule you own assumes one output per input. So at first glance the circle has no derivative to give.
Analogy
Discussion prompt
Explain Why the circle seems out of reach 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 vertical line at almost any x crosses the circle twice. Two outputs for one input means it fails the vertical line test.
Concept
Solve the circle equation for y and you get two half-circles, an upper one and a lower one.
\[ y = \sqrt{25-x^2} \quad \text{(upper)} \qquad y = -\sqrt{25-x^2} \quad \text{(lower)} \]
Each branch is a function, and the chain rule handles each one. It works. It is just more algebra than the problem deserves, and you have to keep track of which branch you are standing on.
Explain it
Discussion prompt
Explain You could split it into branches 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:
Solve the circle equation for y and you get two half-circles, an upper one and a lower one.
Concept
The folium of Descartes:
\[ x^3 + y^3 = 6xy \]
Try to solve that for y. It is a cubic in y, so a formula exists, but it is monstrous and nobody differentiates it.
We need a method that never solves for y in the first place.
Intuition
Pick one point on the circle - say three to the right and four up. Now zoom in on a tiny patch around it.
In that window you see a single smooth arc: one output for each input. Locally, near that point, y really is a function of x.
That is all a derivative ever needs. Differentiation is a local question, so a local function is enough.
Concept
Write the equation as if the unknown function were already sitting inside it, even though you will never find its formula.
\[ x^2 + \left[\,y(x)\,\right]^2 = 25 \]
implicit function — A function of x defined indirectly by an equation instead of by a formula. Near almost every point of a smooth curve such a function exists, even when you cannot write it down.
Concept
Differentiate both sides with respect to x. On any term built out of y, the chain rule fires: differentiate the outside, then multiply by the derivative of the inside - and the inside is y.
\[ \frac{d}{dx}\left[y^2\right] = 2y\,\frac{dy}{dx} \]
That extra factor is implicit differentiation. Everything after it is algebra you already know.
Ranking
Put in order
Put the moves of Warm-up: differentiating y-expressions 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. Power rule on the outside gives three y squared; then multiply by the derivative of the inside, which is dy/dx.
Worked example
Treat y as an unknown function of x and differentiate each expression with respect to x.
Differentiate y cubed
Why: Power rule on the outside gives three y squared; then multiply by the derivative of the inside, which is dy/dx.
\[ \frac{d}{dx}\left[y^3\right] = 3y^2\,\frac{dy}{dx} \]
Differentiate the sine of y
Why: Outside derivative is cosine of y; the inside derivative is still dy/dx.
\[ \frac{d}{dx}\left[\sin y\right] = \cos y\,\frac{dy}{dx} \]
Differentiate e raised to the y
Why: The natural exponential is its own derivative, and the chain rule still attaches the inside derivative.
\[ \frac{d}{dx}\left[e^{y}\right] = e^{y}\,\frac{dy}{dx} \]
Verify by replacing y with x
Why: If y happened to be x itself, then dy/dx becomes dx/dx, which is 1, and every line above collapses to the familiar rule. That consistency is the check.
\[ 3y^2\frac{dy}{dx} \;\longrightarrow\; 3x^2\cdot 1 = \frac{d}{dx}\left[x^3\right] \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Warm-up: differentiating y-expressions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If y happened to be x itself, then dy/dx becomes dx/dx, which is 1, and every line above collapses to the familiar rule. That consistency is the check.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Differentiate, but forget that y depends on x.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Dividing by two gives a relation between x and y, not a slope.
Differentiate, letting the chain rule fire on the y term.
Why: Dividing by two gives a relation between x and y, not a slope.
Trap
Differentiate, but forget that y depends on x.
\[ x^2+y^2=25 \;\longrightarrow\; 2x + 2y = 0 \]
Solve what you got
Why: Dividing by two gives a relation between x and y, not a slope.
\[ y = -x \]
That claims the circle is a straight line. The point three right and four up is on the circle, and four is not the opposite of three - so the conclusion is false and the differentiation was wrong.
Differentiate, letting the chain rule fire on the y term.
\[ x^2+y^2=25 \;\longrightarrow\; 2x + 2y\frac{dy}{dx} = 0 \]
Solve for dy/dx
Why: Isolate the derivative. What comes out is a slope formula, not a curve equation.
\[ \frac{dy}{dx} = -\frac{x}{y} \]
At the point three right and four up the slope is negative three fourths - exactly what the picture shows.
Notation
Annotate
From Trap: differentiating a y term as if it were a constant — read this one piece at a time. What is each part doing?
On: \( x^2+y^2=25 \;\longrightarrow\; 2x + 2y = 0 \)
Section
Part 2
Step zero
Discussion prompt
Example 1: the derivative of the circle — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate both sides with respect to x
Answer:
Worked example
\[ x^2 + y^2 = 25 \]
Find dy/dx.
Differentiate both sides with respect to x
Why: Both sides are functions of x, so both get the same treatment. The right side is a constant.
\[ \frac{d}{dx}\left[x^2+y^2\right] = \frac{d}{dx}\left[25\right] \]
Apply the rules term by term
Why: The x term is ordinary. The y term needs the chain rule, so it carries a dy/dx factor. The constant dies.
\[ 2x + 2y\frac{dy}{dx} = 0 \]
Move the derivative-free term across
Why: Everything without a dy/dx belongs on the other side.
\[ 2y\frac{dy}{dx} = -2x \]
Divide by the coefficient of dy/dx
Why: The twos cancel, leaving the slope in terms of both coordinates.
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Verify against the explicit upper branch
Why: On the top half y is the square root of twenty-five minus x squared. Its chain-rule derivative is negative x over that same square root - and that square root is y. The two answers agree.
\[ \frac{d}{dx}\sqrt{25-x^2} = \frac{-x}{\sqrt{25-x^2}} = -\frac{x}{y} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 1: the derivative of the circle", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: On the top half y is the square root of twenty-five minus x squared. Its chain-rule derivative is negative x over that same square root - and that square root is y. The two answers agree.
Concept
An explicit derivative is a formula in x alone. An implicit derivative usually involves both coordinates - and that is a feature, not a defect.
The circle has two points above most x values, one on top and one on the bottom, and their slopes are opposites. A formula in x alone could not tell them apart.
\[ \left.\frac{dy}{dx}\right|_{(3,4)} = -\frac{3}{4}, \qquad \left.\frac{dy}{dx}\right|_{(3,-4)} = \frac{3}{4} \]
Intuition
Draw the radius from the origin out to the point three right and four up. Its slope is four thirds.
A tangent to a circle is perpendicular to the radius at the point of contact, so its slope must be the negative reciprocal: negative three fourths.
Implicit differentiation reproduced that geometry fact in one line - and it did it for every point of the circle at once.
Concept
A term containing both letters, like x times y, is a product of two functions of x: the plain function x and the unknown function y.
\[ \frac{d}{dx}\left[xy\right] = 1\cdot y + x\cdot\frac{dy}{dx} = y + x\frac{dy}{dx} \]
One piece comes out with no derivative attached, the other with one. Missing that first piece is the second most common implicit error.
Estimation
Predict first
Differentiate each product with respect to x, remembering that y is a function of x.
Commit before you compute: what does Warm-up: differentiating mixed terms 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 the first result by setting y equal to x
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. If y were x, the expression would be x to the fifth, whose derivative is five x to the fourth.
Worked example
Differentiate each product with respect to x, remembering that y is a function of x.
Differentiate x squared times y cubed
Why: Product rule: derivative of the first times the second, plus the first times the derivative of the second. That second derivative needs the chain rule.
\[ \frac{d}{dx}\left[x^2y^3\right] = 2xy^3 + 3x^2y^2\frac{dy}{dx} \]
Differentiate x times the sine of y
Why: Same shape. The derivative of the sine of y is the cosine of y times dy/dx.
\[ \frac{d}{dx}\left[x\sin y\right] = \sin y + x\cos y\,\frac{dy}{dx} \]
Verify the first result by setting y equal to x
Why: If y were x, the expression would be x to the fifth, whose derivative is five x to the fourth. Our formula gives two x to the fourth plus three x to the fourth. They match.
\[ 2x\cdot x^3 + 3x^2\cdot x^2\cdot 1 = 5x^4 = \frac{d}{dx}\left[x^5\right] \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Warm-up: differentiating mixed terms", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: If y were x, the expression would be x to the fifth, whose derivative is five x to the fourth. Our formula gives two x to the fourth plus three x to the fourth. They match.
Fill the middle
Fill in the blanks
From Trap: forgetting the product rule on an x-times-y term — finish the line. Write what belongs on the right of the equals sign before you look.
\frac\frac{-2x}{x+2y} \;\longrightarrow\; -\frac{2}{5}___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The missing piece never reaches the numerator, so the reported slope is too small in size.
Trap
Differentiate, but treat the middle term as though only the y in it were moving.
\[ x^2 + xy + y^2 = 7 \]
\[ 2x + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \]
Solve, then evaluate at the point one right and two up
Why: The missing piece never reaches the numerator, so the reported slope is too small in size.
\[ \frac{dy}{dx} = \frac{-2x}{x+2y} \;\longrightarrow\; -\frac{2}{5} \]
The middle term is a product, so both factors get a turn.
\[ x^2 + xy + y^2 = 7 \]
\[ 2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \]
Solve, then evaluate at the same point
Why: The stray y joins the numerator and changes the answer.
\[ \frac{dy}{dx} = -\frac{2x+y}{x+2y} \;\longrightarrow\; -\frac{4}{5} \]
Missing information
Discussion prompt
Find dy/dx, then its value at the point one right and two up.
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:
One plus two plus four is seven. Never chase a tangent at a point that is not there.
Worked example
\[ x^2 + xy + y^2 = 7 \]
Find dy/dx, then its value at the point one right and two up.
Check that the point is on the curve
Why: One plus two plus four is seven. Never chase a tangent at a point that is not there.
\[ 1^2 + (1)(2) + 2^2 = 7 \quad \checkmark \]
Differentiate every term with respect to x
Why: Ordinary rule on the x term, product rule on the middle term, chain rule on the y term.
\[ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \]
Move every derivative term to one side
Why: Sort the terms into two piles: those carrying dy/dx and those not.
\[ x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y \]
Factor out dy/dx
Why: It is a common factor of both terms, so this is ordinary factoring.
\[ (x+2y)\frac{dy}{dx} = -(2x+y) \]
Divide by what is left
Why: Now there is a single coefficient to divide by, which is what makes the division legal.
\[ \frac{dy}{dx} = -\frac{2x+y}{x+2y} \]
Substitute the point last
Why: Only now do numbers enter. Two plus two over one plus four gives four fifths, with a minus sign.
\[ \left.\frac{dy}{dx}\right|_{(1,2)} = -\frac{2+2}{1+4} = -\frac{4}{5} \]
Verify by putting the point and the slope back into the differentiated equation
Why: Two plus two minus four fifths minus sixteen fifths is four minus four, which is zero - the equation is satisfied, so the slope is consistent with the curve.
\[ 2(1) + 2 + (1)\left(-\tfrac{4}{5}\right) + 2(2)\left(-\tfrac{4}{5}\right) = 4 - \tfrac{4}{5} - \tfrac{16}{5} = 0 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 2: a curve with a mixed term", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One plus two plus four is seven. Never chase a tangent at a point that is not there.
Concept
After you differentiate, every single term either carries a derivative factor or it does not. That splits the equation into two piles, and the rest is one line of algebra.
\[ A(x,y)\,\frac{dy}{dx} + B(x,y) = 0 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{B(x,y)}{A(x,y)} \]
Every implicit problem lands in this shape. Gather, factor, divide - always in that order.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Move the plain terms across, then divide by the first coefficient you happen to see.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Illegal: another dy/dx term is still sitting on the left, so 2y is not the whole coefficient.
Gather every derivative term first, then factor.
Why: Illegal: another dy/dx term is still sitting on the left, so 2y is not the whole coefficient.
Trap
Move the plain terms across, then divide by the first coefficient you happen to see.
\[ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \]
Divide by 2y only
Why: Illegal: another dy/dx term is still sitting on the left, so 2y is not the whole coefficient.
\[ \frac{dy}{dx} = -\frac{2x+y}{2y} \;\longrightarrow\; -\frac{4}{4} = -1 \]
The reported slope at the point one right and two up is negative one. It is wrong.
Gather every derivative term first, then factor.
\[ 2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \]
Factor dy/dx out of both terms
Why: Only after factoring is there a single coefficient, and only then may you divide by it.
\[ (x+2y)\frac{dy}{dx} = -(2x+y) \]
\[ \frac{dy}{dx} = -\frac{2x+y}{x+2y} \;\longrightarrow\; -\frac{4}{5} \]
The true slope there is negative four fifths.
Translation
\( \frac{dy}{dx} = -\frac{2x+y}{x+2y} \;\longrightarrow\; -\frac{4}{5} \)
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.
Constraint
Discussion prompt
Run The implicit differentiation procedure with this step confiscated:
On every term containing y, apply the chain rule and attach a dy/dx factor.
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:
dy/dx factor.dy/dx term to one side, factor it out, and divide by what remains.Pattern
Seven moves, in this order, every time.
dy/dx factor.dy/dx term to one side, factor it out, and divide by what remains.Step seven is the one students break. Substituting numbers before you differentiate destroys the very information you were about to extract.
Picture it
Animation
Shows: When implicit differentiation is the tool — a rendered Manim animation.
Rendered with Manim.
Takeaway: Not a last resort — often the shortest route available.
Check
Work it on paper before you choose.
\[ xy + y^2 = 6 \]
Check your understanding
Find dy/dx for the curve shown above.
Answer: A
Why: Differentiating gives y + x(dy/dx) + 2y(dy/dx) = 0. Collecting the derivative terms gives (x + 2y)(dy/dx) = -y, so dy/dx = -y/(x + 2y).
Section
Part 3
Concept
Point-slope form is unchanged.
\[ y - y_0 = m\,(x - x_0) \]
The problem hands you the point. Implicit differentiation hands you the slope. The only new wrinkle is that the slope formula wants both coordinates, not just x.
Picture it
Animation
Shows: A tangent line still needs a point — a rendered Manim animation.
Rendered with Manim.
Takeaway: Unlike an explicit derivative, one number is not enough.
Picture it
Animation
Shows: A tangent line without solving for y — a rendered Manim animation.
Rendered with Manim.
Takeaway: The point supplies the numbers the implicit derivative needs.
Worked example
\[ x^2 + 4y^2 = 8 \]
Find the tangent line at the point two right and one up.
Check the point is on the curve
Why: Four plus four is eight, so the point lies on the ellipse.
\[ 2^2 + 4(1)^2 = 8 \quad \checkmark \]
Differentiate both sides
Why: The y-squared term carries the chain rule; the constant four just rides along as a coefficient.
\[ 2x + 8y\frac{dy}{dx} = 0 \]
Solve for dy/dx
Why: Isolate and cancel the common factor of two.
\[ \frac{dy}{dx} = -\frac{x}{4y} \]
Evaluate at the point
Why: Substitute both coordinates to turn the formula into a number.
\[ \left.\frac{dy}{dx}\right|_{(2,1)} = -\frac{2}{4} = -\frac{1}{2} \]
Write point-slope form and simplify
Why: Slope one half down, through the given point.
\[ y - 1 = -\tfrac{1}{2}(x-2) \quad\Longrightarrow\quad y = -\tfrac{x}{2} + 2 \]
Verify that the line meets the ellipse exactly once
Why: Substituting the line into the ellipse gives two x squared minus eight x plus sixteen equals eight, which reduces to x minus two, squared, equals zero. A double root is precisely what tangency looks like.
\[ x^2 + 4\left(2-\tfrac{x}{2}\right)^2 = 8 \;\Longrightarrow\; 2x^2 - 8x + 8 = 0 \;\Longrightarrow\; (x-2)^2 = 0 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 3: tangent to an ellipse", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Four plus four is eight, so the point lies on the ellipse.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Put the numbers in first to make everything concrete.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Both sides are plain constants, so both derivatives are zero.
Differentiate first, while x and y are still free to move.
Why: Both sides are plain constants, so both derivatives are zero.
Trap
Put the numbers in first to make everything concrete.
\[ x^2+4y^2=8 \;\longrightarrow\; 2^2 + 4(1)^2 = 8 \]
Now differentiate
Why: Both sides are plain constants, so both derivatives are zero.
\[ \frac{d}{dx}\left[8\right] = \frac{d}{dx}\left[8\right] \;\Longrightarrow\; 0 = 0 \]
True, useless, and not a slope in sight. Freezing the coordinates froze the very motion you were trying to measure.
Differentiate first, while x and y are still free to move.
\[ x^2+4y^2=8 \;\longrightarrow\; 2x + 8y\frac{dy}{dx} = 0 \]
Solve, then substitute
Why: The general formula comes first; the point simply selects one slope out of it.
\[ \frac{dy}{dx} = -\frac{x}{4y} \]
\[ \left.\frac{dy}{dx}\right|_{(2,1)} = -\frac{1}{2} \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The general formula comes first; the point simply selects one slope out of it.
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:
Both sides are plain constants, so both derivatives are zero.
Intuition
A derivative is about a whole neighborhood of the point, not the point alone. It compares the curve here with the curve a hair away.
Substituting numbers first throws that neighborhood away. You are left with a true statement about one point and no information whatsoever about the nearby ones.
Differentiate while the letters are still letters. Numbers go in last.
Sorting
Sort into buckets
These are the pieces of Implicit and Logarithmic Differentiation, out of order. Put each one back under the part of the lesson it belongs to.
Step zero
Discussion prompt
Example 4: the folium of Descartes — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Check the point
Answer:
Worked example
\[ x^3 + y^3 = 6xy \]
Find the slope of the tangent at the point three right and three up.
Check the point
Why: Twenty-seven plus twenty-seven is fifty-four, and six times three times three is also fifty-four.
\[ 3^3 + 3^3 = 54 = 6(3)(3) \quad \checkmark \]
Differentiate both sides
Why: Left side: power rule, with the chain rule on the y term. Right side: product rule, since six x times y is a product of two functions of x.
\[ 3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx} \]
Gather the derivative terms on the left
Why: The dy/dx term on the right has to cross over before you can factor.
\[ 3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2 \]
Factor and divide
Why: Factor dy/dx out on the left, divide, then cancel the common factor of three.
\[ \frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x} = \frac{2y-x^2}{y^2-2x} \]
Evaluate at the point
Why: Six minus nine over nine minus six.
\[ \left.\frac{dy}{dx}\right|_{(3,3)} = \frac{6-9}{9-6} = -1 \]
Verify by substituting the point and the slope into the differentiated equation
Why: Left side: twenty-seven plus twenty-seven times negative one is zero. Right side: eighteen plus eighteen times negative one is zero. Both sides agree, so the slope is consistent.
\[ 3(9) + 3(9)(-1) = 0 \qquad 6(3) + 6(3)(-1) = 0 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 4: the folium of Descartes", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Twenty-seven plus twenty-seven is fifty-four, and six times three times three is also fifty-four.
Elimination
Eliminate the wrong options
What is the slope of the tangent line to that circle at the point where x is negative three and y is four?
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: Implicit differentiation gives dy/dx = -x/y. With x equal to -3 and y equal to 4 that is -(-3)/4 = 3/4. The radius through that point has slope -4/3, and 3/4 is its negative reciprocal, exactly as a circle's tangent must be.
Check
The circle of radius five centered at the origin passes through the point three to the left and four up.
\[ x^2 + y^2 = 25 \]
Check your understanding
What is the slope of the tangent line to that circle at the point where x is negative three and y is four?
Answer: A
Why: Implicit differentiation gives dy/dx = -x/y. With x equal to -3 and y equal to 4 that is -(-3)/4 = 3/4. The radius through that point has slope -4/3, and 3/4 is its negative reciprocal, exactly as a circle's tangent must be.
Section
Part 4
Concept
Once dy/dx is a single fraction, the two special directions are easy to find.
\[ \frac{dy}{dx} = -\frac{2x+y}{x+2y} \]
In every case the candidate must also satisfy the original equation - being on the curve is a separate condition from the slope condition.
Picture it
Animation
Shows: Where the tangent goes vertical — a rendered Manim animation.
Rendered with Manim.
Takeaway: The denominator of the implicit derivative marks the vertical tangents.
Fill the middle
Fill in the blanks
From Example 5: every horizontal tangent on a tilted ellipse — finish the line. Write what belongs on the right of the equals sign before you look.
\frac-\frac{2x+y}{x+2y}___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Same three moves as before: differentiate, gather, factor.
Worked example
\[ x^2 + xy + y^2 = 12 \]
This is an ellipse tilted off the axes. Find every point where its tangent is horizontal.
Differentiate and solve for the slope
Why: Same three moves as before: differentiate, gather, factor. The mixed term uses the product rule.
\[ \frac{dy}{dx} = -\frac{2x+y}{x+2y} \]
Set the numerator equal to zero
Why: A fraction is zero exactly when its top is zero and its bottom is not.
\[ 2x + y = 0 \;\Longrightarrow\; y = -2x \]
Substitute that relation into the original curve
Why: The point must satisfy both conditions at once: lie on the curve and have a zero numerator.
\[ x^2 + x(-2x) + (-2x)^2 = 12 \]
Simplify and solve
Why: One minus two plus four is three, so the whole left side is three x squared.
\[ 3x^2 = 12 \;\Longrightarrow\; x = \pm 2 \]
Report both points
Why: Use the relation y equals negative two x to recover each partner coordinate.
\[ (2,\,-4) \qquad \text{and} \qquad (-2,\,4) \]
Verify the first point on both counts
Why: On the curve: four minus eight plus sixteen is twelve. In the slope formula the numerator is four plus negative four, which is zero, and the denominator is two plus negative eight, which is negative six - nonzero, so the slope truly is zero and not undefined.
\[ 4 - 8 + 16 = 12, \qquad \left.\frac{dy}{dx}\right|_{(2,-4)} = -\frac{0}{-6} = 0 \quad \checkmark \]
Worked example
\[ x^2 + xy + y^2 = 12 \]
Now find the points where the tangent is vertical.
Set the denominator equal to zero
Why: The slope is undefined exactly where the bottom of the fraction vanishes and the top does not.
\[ x + 2y = 0 \;\Longrightarrow\; x = -2y \]
Substitute into the curve
Why: Again the point must satisfy both conditions.
\[ (-2y)^2 + (-2y)y + y^2 = 12 \]
Simplify and solve
Why: Four minus two plus one is three, so this is three y squared.
\[ 3y^2 = 12 \;\Longrightarrow\; y = \pm 2 \]
Report both points
Why: Use x equals negative two y to get each partner coordinate.
\[ (-4,\,2) \qquad \text{and} \qquad (4,\,-2) \]
Verify the first point on both counts
Why: On the curve: sixteen minus eight plus four is twelve. In the slope formula the denominator is negative four plus four, which is zero, while the numerator is negative eight plus two, which is negative six - nonzero, so the tangent is genuinely vertical.
\[ 16 - 8 + 4 = 12, \qquad \left.\frac{dy}{dx}\right|_{(-4,2)} = -\frac{-6}{0} \;\text{(undefined)} \quad \checkmark \]
Intuition
Figure (svg): A tilted ellipse with two marked points at its highest and lowest spots and two more at its far left and far right edges
A horizontal tangent sits at the top or the bottom of the curve: the curve stops rising and starts falling.
A vertical tangent sits at the far left or far right edge: the curve stops moving right and starts moving left. A tilted ellipse has exactly two of each, and they are four different points.
Concept
You already have a formula for the first derivative. Differentiate that formula again with respect to x. It is one more implicit differentiation, nothing new.
Two things to watch: the quotient rule almost always appears, and every dy/dx that pops out along the way must be replaced by the formula you already found.
The answer is finished only when no derivative symbol is left on the right-hand side.
Picture it
Animation
Shows: The second derivative, implicitly — a rendered Manim animation.
Rendered with Manim.
Takeaway: The first derivative reappears inside the second — substitute it away.
Picture it
Animation
Shows: The circle equation differentiated implicitly to a slope formula.
Every y carries a dy/dx with it.
Takeaway: You never solve for y. Differentiate both sides as they stand, attaching dy/dx to every y, then solve for that one symbol at the end.
Hypothesis
Predict first
Example 7: the circle's second derivative 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 the slope formula with the quotient rule
Why: Top is x, bottom is y. The derivative of the top is 1; the derivative of the bottom is dy/dx, not 1, because y is a function of x.
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
\[ x^2+y^2=25, \qquad \frac{dy}{dx} = -\frac{x}{y} \]
Find the second derivative.
Differentiate the slope formula with the quotient rule
Why: Top is x, bottom is y. The derivative of the top is 1; the derivative of the bottom is dy/dx, not 1, because y is a function of x.
\[ \frac{d^2y}{dx^2} = -\frac{(1)(y) - x\frac{dy}{dx}}{y^2} \]
Substitute the first derivative you already found
Why: Replace dy/dx by negative x over y. This is the step that removes every derivative from the right-hand side.
\[ = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} \]
Clear the inner fraction
Why: Multiply the numerator and the denominator of the big fraction by y.
\[ = -\frac{y^2 + x^2}{y^3} \]
Use the original equation to simplify
Why: On this curve x squared plus y squared is twenty-five, so the whole numerator collapses to a number.
\[ \frac{d^2y}{dx^2} = -\frac{25}{y^3} \]
Verify against the explicit upper branch at the point three right and four up
Why: Differentiating the top half twice gives negative one quarter minus nine sixty-fourths, which is negative twenty-five sixty-fourths - and our formula gives negative twenty-five over four cubed, the same number. The sign is negative, matching the fact that the top of a circle is concave down.
\[ -\left(25-x^2\right)^{-1/2} - x^2\left(25-x^2\right)^{-3/2}\Big|_{x=3} = -\tfrac{1}{4}-\tfrac{9}{64} = -\tfrac{25}{64} = -\frac{25}{4^3} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 7: the circle's second derivative", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating the top half twice gives negative one quarter minus nine sixty-fourths, which is negative twenty-five sixty-fourths - and our formula gives negative twenty-five over four cubed, the same number. The sign is negative, matching the fact that the top of a circle is concave down.
Trap
Differentiate again, but treat the y in the denominator as a constant.
\[ \frac{dy}{dx} = -\frac{x}{y} \;\longrightarrow\; \frac{d^2y}{dx^2} = -\frac{1}{y} \]
Evaluate at the point three right and four up
Why: This reports a concavity of negative one quarter.
\[ -\frac{1}{4} = -0.25 \]
The number is too small in size, because the moving denominator was never allowed to contribute.
The denominator is a function of x, so the quotient rule applies and dy/dx reappears - then gets substituted away.
\[ \frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{25}{y^3} \]
Evaluate at the same point
Why: Negative twenty-five over sixty-four, which is about negative zero point three nine.
\[ -\frac{25}{64} \approx -0.39 \]
Over half again as steep a bend as the wrong answer claimed.
Prediction
Predict first
At which points does this curve have a vertical tangent line?
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: (-4, 2) and (4, -2)
Why: A vertical tangent needs the denominator x + 2y to be zero with the numerator nonzero. Setting x = -2y in the curve gives 3y squared = 12, so y = 2 or y = -2, producing the points (-4, 2) and (4, -2).
Check
Use the slope formula you built for the tilted ellipse earlier in this deck.
\[ x^2+xy+y^2=12, \qquad \frac{dy}{dx} = -\frac{2x+y}{x+2y} \]
Check your understanding
At which points does this curve have a vertical tangent line?
Answer: A
Why: A vertical tangent needs the denominator x + 2y to be zero with the numerator nonzero. Setting x = -2y in the curve gives 3y squared = 12, so y = 2 or y = -2, producing the points (-4, 2) and (4, -2).
Section
Part 5
Concept
Suppose y is the natural logarithm of x. By the very meaning of a logarithm, that says e raised to the y equals x.
\[ y = \ln x \iff e^{y} = x \]
The second form has no logarithm in it at all - and it is exactly the kind of equation implicit differentiation eats for breakfast.
Picture it
Animation
Shows: An inverse is an implicit equation in disguise — a rendered Manim animation.
Rendered with Manim.
Takeaway: Every inverse-function derivative is this one line.
Picture it
Animation
Shows: The chain rule producing the dy/dx factor on a y term.
It is the chain rule, not a new rule.
Takeaway: The dy/dx is not a convention — it is the chain rule applied to a function of x that happens to be called y.
Intuition
The graph of an inverse function is the graph of the original reflected across the diagonal line where the two coordinates are equal.
Reflecting swaps the run and the rise of every little step along the curve. A step that went two right and one up becomes a step that goes one right and two up.
Swapping rise and run inverts the slope. So the inverse's derivative should be the reciprocal of the original's derivative - at the matching point. That is the whole theorem, before any algebra.
Ranking
Put in order
Put the moves of Example 8: deriving the derivative of the natural log 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 of the logarithm, not a manipulation, so nothing has been assumed.
Worked example
\[ y = \ln x, \qquad x > 0 \]
Find dy/dx without assuming the answer.
Rewrite as an exponential equation
Why: This is the definition of the logarithm, not a manipulation, so nothing has been assumed.
\[ e^{y} = x \]
Differentiate both sides with respect to x
Why: The left side needs the chain rule, because the exponent is the unknown function y.
\[ e^{y}\frac{dy}{dx} = 1 \]
Solve for dy/dx
Why: Divide by the coefficient, which is the exponential itself.
\[ \frac{dy}{dx} = \frac{1}{e^{y}} \]
Translate back into x
Why: The equation we started from says e raised to the y is x, so the denominator is just x.
\[ \frac{dy}{dx} = \frac{1}{x} \]
Verify by differentiating the identity e raised to the log of x
Why: That composition simplifies to x, so its derivative has to be 1. The chain rule with our answer gives x times one over x, which is indeed 1.
\[ \frac{d}{dx}\left[e^{\ln x}\right] = e^{\ln x}\cdot\frac{1}{x} = x\cdot\frac{1}{x} = 1 = \frac{d}{dx}\left[x\right] \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 8: deriving the derivative of the natural log", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: That composition simplifies to x, so its derivative has to be 1. The chain rule with our answer gives x times one over x, which is indeed 1.
Estimation
Predict first
Same trick: rewrite it as the equation that defines it.
Commit before you compute: what does Example 9: the derivative of arcsine 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 a known angle
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 one half the angle is thirty degrees, whose cosine is root three over two, so the slope should be two over root three, about 1.155.
Worked example
\[ y = \arcsin x, \qquad -\frac{\pi}{2} \le y \le \frac{\pi}{2} \]
Same trick: rewrite it as the equation that defines it.
Rewrite without the inverse
Why: Arcsine of x is by definition the angle in that range whose sine is x.
\[ \sin y = x \]
Differentiate both sides
Why: Chain rule on the left: derivative of sine is cosine, times the derivative of the inside.
\[ \cos y \,\frac{dy}{dx} = 1 \]
Solve for dy/dx
Why: Divide by the cosine factor.
\[ \frac{dy}{dx} = \frac{1}{\cos y} \]
Rewrite the cosine in terms of x
Why: Pythagorean identity, and on the arcsine range the cosine is never negative, so the positive square root is the correct one.
\[ \cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2} \]
State the result
Why: Valid for x strictly between negative one and one.
\[ \frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^2}} \]
Verify at a known angle
Why: At x equal to one half the angle is thirty degrees, whose cosine is root three over two, so the slope should be two over root three, about 1.155. The formula gives one over the square root of three quarters, which is the same number. It is positive, matching an increasing function, and it blows up as x approaches one, matching the vertical tangent there.
\[ \frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt3} \approx 1.155 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 9: the derivative of arcsine", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At x equal to one half the angle is thirty degrees, whose cosine is root three over two, so the slope should be two over root three, about 1.155. The formula gives one over the square root of three quarters, which is the same number. It is positive, matching an increasing function, and it blows up as x approaches one, matching the vertical tangent there.
Concept
Now do it in general. If g undoes f, then feeding g of x into f gives back x.
\[ f\bigl(g(x)\bigr) = x \]
Differentiate both sides with the chain rule and solve.
\[ f'\bigl(g(x)\bigr)\,g'(x) = 1 \quad\Longrightarrow\quad g'(x) = \frac{1}{f'\bigl(g(x)\bigr)} \]
The slope of the inverse is the reciprocal of the slope of the original - but evaluated at the matching input, not at the same one. That inner detail is where the mistakes live.
Step zero
Discussion prompt
Example 10: a derivative with no formula in sight — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Find the matching point
Answer:
Worked example
\[ f(x) = x^3 + x + 1 \]
This function is increasing everywhere, so it has an inverse g. You cannot write a clean formula for g. Find the slope of g at the input three.
Find the matching point
Why: Solve f of x equals three by inspection: one plus one plus one is three, so g of three is one.
\[ f(1) = 1+1+1 = 3 \;\Longrightarrow\; g(3) = 1 \]
Differentiate f
Why: Ordinary power and sum rules - the hard function never needs to be differentiated.
\[ f'(x) = 3x^2 + 1 \]
Evaluate f prime at the matching input
Why: At one, not at three. This is the step that gets skipped.
\[ f'(1) = 3+1 = 4 \]
Take the reciprocal
Why: The inverse function theorem.
\[ g'(3) = \frac{1}{f'(1)} = \frac{1}{4} \]
Verify numerically
Why: Step x from 1 to 1.01: f climbs from 3 to 3.040301. The inverse reverses those roles, so its average slope over that step is 0.01 divided by 0.040301, about 0.248 - closing in on one quarter as the step shrinks.
\[ \frac{1.01 - 1}{3.040301 - 3} = \frac{0.01}{0.040301} \approx 0.2481 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 10: a derivative with no formula in sight", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Step x from 1 to 1.01: f climbs from 3 to 3.040301. The inverse reverses those roles, so its average slope over that step is 0.01 divided by 0.040301, about 0.248 - closing in on one quarter as the step shrinks.
Commit first
Predict first
What is the value of the derivative of g at the input 3?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: 1/5
Why: Since f(1) = 1 + 2 = 3, the matching point is g(3) = 1. Then f'(x) = 3x^2 + 2 gives f'(1) = 5, and the inverse function theorem says the derivative of g at 3 is 1 divided by 5.
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
Let g be the inverse of the function below, which is increasing everywhere.
\[ f(x) = x^3 + 2x \]
Check your understanding
What is the value of the derivative of g at the input 3?
Answer: A
Why: Since f(1) = 1 + 2 = 3, the matching point is g(3) = 1. Then f'(x) = 3x^2 + 2 gives f'(1) = 5, and the inverse function theorem says the derivative of g at 3 is 1 divided by 5.
Section
Part 6
Concept
Look at this and imagine the product rule inside the quotient rule inside the chain rule.
\[ y = \frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5} \]
It is doable with the rules you already own. It is also four lines of bookkeeping in which a single dropped sign ends the problem.
Intuition
A logarithm converts a product into a sum, a quotient into a difference, and an exponent into a plain multiplier out front.
Sums and multipliers are the easiest things in all of calculus to differentiate. So: take the log, differentiate the easy thing, then convert back.
That is the whole idea. You are not finding a new derivative rule - you are trading a hard problem for an easy one you already know how to finish.
Explain it
Discussion prompt
Explain Logs turn multiplication into addition 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:
A logarithm converts a product into a sum, a quotient into a difference, and an exponent into a plain multiplier out front.
Picture it
Animation
Shows: Logs turn products into sums — a rendered Manim animation.
Rendered with Manim.
Takeaway: A four-rule nightmare becomes a sum of easy pieces.
Concept
\[ \ln(ab) = \ln a + \ln b \]
\[ \ln\!\left(\frac{a}{b}\right) = \ln a - \ln b \]
\[ \ln\!\left(a^{\,n}\right) = n\ln a \]
Every logarithmic differentiation problem is these three laws, run left to right until nothing complicated is left inside a logarithm, followed by the chain rule.
Analogy
Discussion prompt
Explain The three log laws you will use 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:
Every logarithmic differentiation problem is these three laws, run left to right until nothing complicated is left inside a logarithm, followed by the chain rule.
Concept
Differentiate the natural log of y with respect to x. The chain rule produces something worth naming.
\[ \frac{d}{dx}\left[\ln y\right] = \frac{1}{y}\cdot\frac{dy}{dx} = \frac{y'}{y} \]
logarithmic derivative — The quantity y prime divided by y: the relative, or fractional, rate of change of y. Multiply it by y to recover the ordinary derivative.
So after differentiating, the last algebra step is always to multiply back by y.
Definition probe
Sort into buckets
Every line below is part of the definition of implicit function or of logarithmic derivative — one or the other, never both. Put each where it belongs.
Picture it
Animation
Shows: A variable-base variable-exponent expression differentiated via logs.
Neither the power rule nor the exponential rule reaches this.
Takeaway: When the base and the exponent both vary, no single rule applies. Taking logs first converts the whole thing into a product you can differentiate.
Ranking
Put in order
These are the steps of The logarithmic differentiation procedure, scrambled. Put them back in order before the next slide shows you.
y' divided by y.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
Reach for it when the expression is a tangle of products, quotients, and powers - or when a variable sits in the base and in the exponent.
y' divided by y.Step four is the one people forget, and forgetting it leaves the answer off by an entire factor.
Edge cases
Discussion prompt
The logarithmic differentiation procedure 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:
Reach for it when the expression is a tangle of products, quotients, and powers - or when a variable sits in the base and in the exponent.
Picture it
Animation
Shows: When to reach for logarithmic differentiation — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two signals, one technique.
Fill the middle
Fill in the blanks
From Example 11: a case you can double-check — finish the line. Write what belongs on the right of the equals sign before you look.
\frac\frac{2}{x+1} + \frac{3}{x+2}___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Legal because the expression is positive on the region we care about; the log is a one-to-one function, so no information is lost.
Worked example
\[ y = (x+1)^2(x+2)^3 \]
Small enough that the product rule would handle it - which is exactly why it makes a good test case.
Take the natural log of both sides
Why: Legal because the expression is positive on the region we care about; the log is a one-to-one function, so no information is lost.
\[ \ln y = \ln\left[(x+1)^2(x+2)^3\right] \]
Expand with the log laws
Why: The product becomes a sum, and each exponent drops down in front as a multiplier.
\[ \ln y = 2\ln(x+1) + 3\ln(x+2) \]
Differentiate both sides
Why: The left side is the logarithmic derivative. On the right, each log gives the inside's derivative over the inside, and here each inside has derivative 1.
\[ \frac{y'}{y} = \frac{2}{x+1} + \frac{3}{x+2} \]
Multiply back by y
Why: Then replace y by the original expression so the answer is in x alone.
\[ y' = (x+1)^2(x+2)^3\left[\frac{2}{x+1} + \frac{3}{x+2}\right] \]
Distribute and cancel
Why: One factor of x plus one cancels in the first term, one factor of x plus two in the second.
\[ y' = 2(x+1)(x+2)^3 + 3(x+1)^2(x+2)^2 \]
Verify with the product rule
Why: Derivative of the first factor times the second, plus the first times the derivative of the second. Two identical terms, in the same order - the two methods agree exactly.
\[ \underbrace{2(x+1)}_{\text{first}'}(x+2)^3 + (x+1)^2\underbrace{3(x+2)^2}_{\text{second}'} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 11: a case you can double-check", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Derivative of the first factor times the second, plus the first times the derivative of the second. Two identical terms, in the same order - the two methods agree exactly.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Differentiate, then report the right-hand side as the derivative.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Two over one plus three over two is three point five.
Multiply both sides by y first, then substitute the original expression.
Why: Two over one plus three over two is three point five.
Trap
Differentiate, then report the right-hand side as the derivative.
\[ \frac{y'}{y} = \frac{2}{x+1} + \frac{3}{x+2} \;\longrightarrow\; y' \stackrel{?}{=} \frac{2}{x+1} + \frac{3}{x+2} \]
Test it at the input zero
Why: Two over one plus three over two is three point five.
\[ \frac{2}{1} + \frac{3}{2} = 3.5 \]
The left side was never y prime. It was y prime divided by y, so the reported answer is missing an entire factor.
Multiply both sides by y first, then substitute the original expression.
\[ y' = y\left[\frac{2}{x+1} + \frac{3}{x+2}\right] \]
Test it at the same input
Why: At zero the function value is one times eight, which is eight, so the derivative is eight times three point five.
\[ y(0) = (1)^2(2)^3 = 8, \qquad y'(0) = 8(3.5) = 28 \]
Twenty-eight, not three point five - the two answers are a factor of eight apart.
Notation
Annotate
From Trap: forgetting to multiply back by y — read this one piece at a time. What is each part doing?
On: \( y(0) = (1)^2(2)^3 = 8, \qquad y'(0) = 8(3.5) = 28 \)
Missing information
Discussion prompt
A root of a quotient: chain rule outside, quotient rule inside. Logs flatten it.
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 square root is the one-half power, so it comes down as a one-half multiplier, and the quotient becomes a difference.
Worked example
\[ y = \sqrt{\frac{x-1}{x+1}}, \qquad x > 1 \]
A root of a quotient: chain rule outside, quotient rule inside. Logs flatten it.
Take the log and expand
Why: The square root is the one-half power, so it comes down as a one-half multiplier, and the quotient becomes a difference.
\[ \ln y = \tfrac{1}{2}\ln(x-1) - \tfrac{1}{2}\ln(x+1) \]
Differentiate both sides
Why: Each log contributes one over its inside, and both insides have derivative 1.
\[ \frac{y'}{y} = \frac{1}{2(x-1)} - \frac{1}{2(x+1)} \]
Combine over a common denominator
Why: The x terms cancel in the numerator, leaving a constant on top and a difference of squares below.
\[ \frac{y'}{y} = \frac{(x+1)-(x-1)}{2(x-1)(x+1)} = \frac{2}{2\left(x^2-1\right)} = \frac{1}{x^2-1} \]
Multiply back by y and simplify
Why: Writing the root as powers lets the factors combine: a one-half power over a first power on the bottom becomes a negative one-half power.
\[ y' = \frac{1}{x^2-1}\sqrt{\frac{x-1}{x+1}} = \frac{1}{\sqrt{x-1}\,(x+1)^{3/2}} \]
Verify with the chain and quotient rules directly
Why: One half of the reciprocal root, times the quotient-rule derivative two over x plus one squared, simplifies to the same expression - so both routes land on the same answer.
\[ \tfrac{1}{2}\left(\frac{x+1}{x-1}\right)^{1/2}\cdot\frac{2}{(x+1)^2} = \frac{1}{\sqrt{x-1}\,(x+1)^{3/2}} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 12: a square root of a quotient", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: One half of the reciprocal root, times the quotient-rule derivative two over x plus one squared, simplifies to the same expression - so both routes land on the same answer.
Estimation
Predict first
The one that looked hopeless. Take the log.
Commit before you compute: what does Example 13: the tangle from the start of this part 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 an end-behavior check
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. For very large x the function behaves like x to the seven fourths over 243 x to the fifth, that is x to the negative thirteen fourths, whose logarithmic derivative is negative thirteen over four x.
Worked example
\[ y = \frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}, \qquad x > 0 \]
The one that looked hopeless. Take the log.
Take the log and expand completely
Why: Three exponents come down in front, the product becomes a sum, and the quotient becomes a difference.
\[ \ln y = \tfrac{3}{4}\ln x + \tfrac{1}{2}\ln\left(x^2+1\right) - 5\ln(3x+2) \]
Differentiate term by term
Why: Each log gives the inside's derivative over the inside. That chain rule inside the log is where the 2x and the 3 come from; the one half then eats the 2, and 5 times 3 makes the 15.
\[ \frac{y'}{y} = \frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2} \]
Multiply back by y
Why: The answer is the original expression times the bracket. There is no reason to expand it - this form is the usable one.
\[ y' = \frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}\left[\frac{3}{4x} + \frac{x}{x^2+1} - \frac{15}{3x+2}\right] \]
Read one value off it
Why: At the input one the bracket is three quarters plus one half minus three, which is negative seven quarters, and the function value is root two over 3125.
\[ y'(1) = \frac{\sqrt2}{3125}\left(-\frac{7}{4}\right) = -\frac{7\sqrt2}{12500} \]
Verify with an end-behavior check
Why: For very large x the function behaves like x to the seven fourths over 243 x to the fifth, that is x to the negative thirteen fourths, whose logarithmic derivative is negative thirteen over four x. Our bracket for large x is three over four x, plus one over x, minus five over x - which is exactly negative thirteen over four x.
\[ \frac{3}{4x} + \frac{1}{x} - \frac{5}{x} = \frac{3-16}{4x} = -\frac{13}{4x} \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 13: the tangle from the start of this part", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For very large x the function behaves like x to the seven fourths over 243 x to the fifth, that is x to the negative thirteen fourths, whose logarithmic derivative is negative thirteen over four x. Our bracket for large x is three over four x, plus one over x, minus five over x - which is exactly negative thirteen over four x.
Concept
A logarithm only accepts positive inputs, so strictly the first step should take the log of the absolute value of both sides.
\[ \ln|y| = \ln|x+1|^2 + \ln|x+2|^3 \;\Longrightarrow\; \frac{d}{dx}\ln|y| = \frac{y'}{y} \]
The good news: the derivative of the log of the absolute value is the same as the derivative of the plain log. So the bars change nothing in the computation - they only widen the set of inputs the answer is valid for.
Counterexample
Discussion prompt
A logarithm only accepts positive inputs, so strictly the first step should take the log of the absolute value of both sides.
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.
Prediction
Predict first
Which expression equals y prime divided by y?
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: 6x/(x squared + 1) + 5/(x - 4)
Why: Taking logs gives ln y = 3 ln(x squared + 1) + 5 ln(x - 4). Differentiating gives 3 times 2x/(x squared + 1) plus 5 times 1/(x - 4), which is 6x/(x squared + 1) + 5/(x - 4).
Check
Take the natural log of both sides, expand, and differentiate. Do not multiply back yet.
\[ y = \left(x^2+1\right)^3(x-4)^5 \]
Check your understanding
Which expression equals y prime divided by y?
Answer: A
Why: Taking logs gives ln y = 3 ln(x squared + 1) + 5 ln(x - 4). Differentiating gives 3 times 2x/(x squared + 1) plus 5 times 1/(x - 4), which is 6x/(x squared + 1) + 5/(x - 4).
Section
Part 7
Concept
Sort every power expression by where the variable is. Three of the four shapes already have a rule.
| Shape | Example | Which rule |
|---|---|---|
| constant base, constant exponent | five cubed | it is a number - derivative zero |
| variable base, constant exponent | x cubed | power rule |
| constant base, variable exponent | two to the x | exponential rule |
| variable base, variable exponent | x to the x | no rule covers it |
That last row is the gap. Logarithmic differentiation is the only elementary way across it.
Comparison
Comparison matrix
From Four shapes, three rules, one gap: refill the Example column from what you know. The rest of the table is as it appeared.
| Shape | Example | Which rule |
|---|---|---|
| constant base, constant exponent | five cubed | it is a number - derivative zero |
| variable base, constant exponent | x cubed | power rule |
| constant base, variable exponent | two to the x | exponential rule |
| variable base, variable exponent | x to the x | no rule covers it |
Step zero
Discussion prompt
Example 14: the derivative of x to the x — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Take the natural log of both sides
Answer:
Worked example
\[ y = x^{x}, \qquad x > 0 \]
The variable is in both places, so take the log.
Take the natural log of both sides
Why: The exponent comes down in front, which is the whole point: the variable exponent becomes an ordinary multiplier.
\[ \ln y = x\ln x \]
Differentiate both sides
Why: Left side is the logarithmic derivative. Right side is a product, so use the product rule: one times the log, plus x times one over x.
\[ \frac{y'}{y} = 1\cdot\ln x + x\cdot\frac{1}{x} = \ln x + 1 \]
Multiply back by y
Why: And replace y by the original expression.
\[ y' = x^{x}\left(\ln x + 1\right) \]
Verify by the exponential rewrite
Why: Any positive base to a power equals e raised to the power times the log of the base. Differentiating that form with the chain rule reproduces the same answer by a completely independent route.
\[ \frac{d}{dx}\,e^{x\ln x} = e^{x\ln x}\left(\ln x + 1\right) = x^{x}\left(\ln x+1\right) \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 14: the derivative of x to the x", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Any positive base to a power equals e raised to the power times the log of the base. Differentiating that form with the chain rule reproduces the same answer by a completely independent route.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Attempt one: use the power rule, bringing the exponent down in front.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The power rule needs a constant exponent; the exponential rule needs a constant base.
Neither rule applies, so take the log and let the product rule do the work.
Why: The power rule needs a constant exponent; the exponential rule needs a constant base. Neither hypothesis holds here, so neither conclusion is trustworthy.
Trap
Attempt one: use the power rule, bringing the exponent down in front.
\[ y' \stackrel{?}{=} x\cdot x^{x-1} = x^{x} \;\longrightarrow\; y'(2) = 4 \]
Attempt two: use the exponential rule, multiplying by the log of the base.
\[ y' \stackrel{?}{=} x^{x}\ln x \;\longrightarrow\; y'(2) = 4\ln 2 \approx 2.77 \]
Notice that both answers are too small
Why: The power rule needs a constant exponent; the exponential rule needs a constant base. Neither hypothesis holds here, so neither conclusion is trustworthy.
Neither rule applies, so take the log and let the product rule do the work.
\[ \ln y = x\ln x \;\Longrightarrow\; \frac{y'}{y} = \ln x + 1 \;\Longrightarrow\; y' = x^{x}(\ln x + 1) \]
Evaluate at the input two
Why: Four times the quantity log of two plus one, which is about 6.77 - noticeably larger than either wrong answer.
\[ y'(2) = 4\left(\ln 2 + 1\right) \approx 6.7726 \]
Settle it numerically
Why: Stepping x from 2 to 2.001 moves the function from 4 to about 4.006780, a slope of roughly 6.78. Only the logarithmic answer is anywhere near it.
\[ \frac{4.006780 - 4}{0.001} \approx 6.78 \]
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.
Ranking
Put in order
Put the moves of Example 15: a trigonometric exponent 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 whole exponent, sine and all, drops in front as a multiplier.
Worked example
\[ y = x^{\sin x}, \qquad x > 0 \]
Variable base, variable exponent again - same move.
Take the natural log
Why: The whole exponent, sine and all, drops in front as a multiplier.
\[ \ln y = \sin x\,\ln x \]
Differentiate with the product rule
Why: Derivative of sine times the log, plus sine times the derivative of the log.
\[ \frac{y'}{y} = \cos x\,\ln x + \frac{\sin x}{x} \]
Multiply back by y
Why: And substitute the original expression for y.
\[ y' = x^{\sin x}\left(\cos x\,\ln x + \frac{\sin x}{x}\right) \]
Verify at a quarter turn
Why: There the sine is 1 and the cosine is 0, so the function value is the angle itself and the slope is that angle times its own reciprocal, which is exactly 1. Rewriting the function as e raised to sine of x times log of x and using the chain rule gives the identical formula, confirming it a second way.
\[ y'\!\left(\tfrac{\pi}{2}\right) = \tfrac{\pi}{2}\left(0 + \tfrac{1}{\pi/2}\right) = 1 \quad \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Example 15: a trigonometric exponent", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: There the sine is 1 and the cosine is 0, so the function value is the angle itself and the slope is that angle times its own reciprocal, which is exactly 1. Rewriting the function as e raised to sine of x times log of x and using the chain rule gives the identical formula, confirming it a second way.
Check
Both the base and the exponent move. Decide what the derivative is.
\[ y = x^{x}, \qquad x>0 \]
Check your understanding
What is dy/dx for the function shown?
Answer: A
Why: Taking logs gives ln y = x ln x. The product rule gives y'/y = ln x + 1, and multiplying back by y gives (x to the x)(ln x + 1). At x = 2 that is about 6.77, matching a numerical slope of about 6.78.
Pattern
Before you differentiate anything, ask what shape the problem is in.
| If the problem looks like... | Reach for... |
|---|---|
| y is already alone on one side | the ordinary differentiation rules |
| x and y are tangled together in one equation | implicit differentiation |
| a long product or quotient of powers and roots | logarithmic differentiation |
| the variable is in the base and in the exponent | logarithmic differentiation - nothing else works |
| a function described only as an inverse | implicit differentiation or the inverse function theorem |
The last two rows overlap on purpose: logarithmic differentiation is implicit differentiation, applied after taking the log of both sides.
Comparison
Comparison matrix
From Choosing the right tool: refill the Reach for... column from what you know. The rest of the table is as it appeared.
| If the problem looks like... | Reach for... |
|---|---|
| y is already alone on one side | the ordinary differentiation rules |
| x and y are tangled together in one equation | implicit differentiation |
| a long product or quotient of powers and roots | logarithmic differentiation |
| the variable is in the base and in the exponent | logarithmic differentiation - nothing else works |
| a function described only as an inverse | implicit differentiation or the inverse function theorem |
Elimination
Eliminate the wrong options
Which method will actually differentiate this function?
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 variable appears in the base and in the exponent, so no power or exponential rule applies. Taking logs gives ln y = x times ln(ln x), which the product and chain rules handle, and then you multiply back by y.
Check
You are asked to differentiate this. Decide the method before touching a pencil.
\[ y = \left(\ln x\right)^{x}, \qquad x > 1 \]
Check your understanding
Which method will actually differentiate this function?
Answer: A
Why: The variable appears in the base and in the exponent, so no power or exponential rule applies. Taking logs gives ln y = x times ln(ln x), which the product and chain rules handle, and then you multiply back by y.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Curves That Are Not Functions · Implicit Differentiation in Action · Tangent Lines on Implicit Curves · Horizontal, Vertical, and Second Derivatives · Derivatives of Inverse Functions · Logarithmic Differentiation. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
One idea carried this entire deck: y is secretly a function of x, so the chain rule fires on every y term and leaves a dy/dx behind.
dy/dx on every y term.dy/dx away.| Symptom | Move |
|---|---|
| a y term | chain rule - attach dy/dx |
| an x-times-y term | product rule first, chain rule inside |
| two or more dy/dx terms | gather them, then factor |
| a numerical answer wanted | substitute the point only at the end |
| a pile of factors and powers | take the natural log of both sides |
| variable base and variable exponent | take the natural log - there is no other rule |
Next up: related rates, where the same chain rule fires on a time variable instead of on x.
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