AP Calculus AB · September 25, 2026

Implicit Differentiation

Differentiate a relationship even when y is not isolated. Apply the chain rule to dependent variables, solve for the desired derivative, and connect the method to changing geometric quantities.

38-minute edited lesson11 chaptersCaptions + corrected transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Implicit Differentiation 5:2302A Circle Written Implicitly 8:3903Differentiate Both Sides 12:2304Tangent Slopes on the Circle 16:0105The Explicit Method Compared 17:0206Differentiate with Respect to Any Variable 20:0007Combine Product and Chain Rules 25:4608A Cylinder and Changing Variables 28:1909A Preview of Related Rates 31:3310Differentiate with Respect to Height 33:5411Combine Quotient, Product, and Chain Rules

Video not loading? Watch directly on YouTube.

02 · Understand

Learning targets

Differentiate both sides of an equation with respect to a chosen variable.
Attach a chain-rule factor whenever a variable depends on the differentiation variable.
Solve an implicit derivative equation for $dy/dx$ or another requested rate.
Use a point on the original curve to evaluate a tangent slope.
Interpret formulas that connect several changing quantities as related rates.

03 · Connect

The implicit method

Leibniz notation makes the dependency visible.

01Differentiate both sides: $x^2+y^2=9$ $2x+2y\frac{dy}{dx}=0$ The derivative of $y^2$ is $2y\,dy/dx$ because y depends on x.
02Solve for the derivative: $\frac{dy}{dx}=-\frac{x}{y}$ Treat the derivative expression as an algebraic quantity: collect it, factor it, and isolate it.
03Evaluate on the curve: $(2,\sqrt5),\ (2,-\sqrt5)$ $m=-\frac{2}{\sqrt5},\quad m=\frac{2}{\sqrt5}$ A single x-coordinate can correspond to two points, so the same curve can have two tangent slopes there.
04Choose another variable: $2x\frac{dx}{dz}+2y\frac{dy}{dz}=0$ $\frac{dy}{dz}=-\frac{x}{y}\frac{dx}{dz}$ When differentiating with respect to z, both x and y receive derivative factors.
05Connect changing quantities: $\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$ This derivative equation relates how volume, radius, and height change over time.

04 · Apply

Worked examples

Examples from the board, reconstructed with accessible typeset mathematics.

Core method

Differentiate a circle implicitly

x^2+y^2=9

Differentiate both sides with respect to x. Treat y as a function of x, so the derivative of y squared includes dy/dx.

Collect the derivative term and solve algebraically for dy/dx.

\frac{dy}{dx}=-\frac{x}{y}

Tangent slopes

Two points share the same x-coordinate

(2,\sqrt5),\ (2,-\sqrt5)

Substitute x=2 into the original curve to find both possible y-values.

Evaluate -x/y at each point. The upper and lower semicircles have opposite tangent slopes.

m=-\frac{2}{\sqrt5},\quad m=\frac{2}{\sqrt5}

Any variable

Differentiate the same curve with respect to z

2x\frac{dx}{dz}+2y\frac{dy}{dz}=0

Now both x and y may depend on z, so both power-rule derivatives need chain-rule factors.

Solve for dy/dz; dx/dz appropriately remains in the answer.

\frac{dy}{dz}=-\frac{x}{y}\frac{dx}{dz}

Combine rules

A product on an implicit curve

\sin x\,e^y=y^3

Use the product rule on sine x times e to the y. Differentiating e to the y introduces dy/dx.

Differentiate y cubed, collect all dy/dx terms, factor, and divide.

\frac{dy}{dx}=\frac{e^y\cos x}{3y^2-e^y\sin x}

Related rates

Differentiate a cylinder with respect to time

V=\pi r^2h

Volume, radius, and height may all change with time.

Use the product rule on r squared times h and attach the appropriate time derivative to each changing variable.

\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}

Multiple rules

A quotient equals a product

\frac{x^4}{\sqrt y}=x^3y

Differentiate both sides with respect to n.

The left side uses the quotient and chain rules; the right side uses the product and chain rules.

\frac{4x^3\sqrt y\,\frac{dx}{dn}-\frac{x^4}{2\sqrt y}\frac{dy}{dn}}{y}=3x^2y\frac{dx}{dn}+x^3\frac{dy}{dn}

05 · Avoid

Common mistakes

Dropping the chain-rule factor.

$d(y^2)/dx=2y\,dy/dx$ —not merely $2y$ .

Solving before differentiating.

Implicit differentiation often avoids square roots, plus-or-minus branches, and unnecessary algebra.

Using the point too early.

Find a general derivative first; then substitute the coordinates from the original curve.

Forgetting multiple y-values.

At $x=2$ on $x^2+y^2=9$ , both $y=\sqrt5$ and $y=-\sqrt5$ matter.

Changing notation inconsistently.

If differentiating with respect to t, use $dx/dt$ , $dy/dt$ , and every other time rate consistently.

Treating derivative symbols as ordinary fractions.

Leibniz notation records a rate and its reference variable; it supports reasoning but is not ordinary fraction arithmetic.

06 · Check yourself

Try it first

These are new problems—not repeats of the worked examples. Reveal an answer only after completing your own derivative.

1For

x^2+y^2=25

, find the tangent slope at

(3,4)

$\left.\frac{dy}{dx}\right|_{(3,4)}=-\frac34$

2Find

dy/dx

x^2+xy+y^2=7

$\frac{dy}{dx}=-\frac{2x+y}{x+2y}$

3Find

dy/dx

e^y\sin x=2

$\frac{dy}{dx}=-\cot x$ After differentiating, divide by $e^y\sin x$ and simplify.

4A cylinder has

r=3

h=5

dr/dt=0.2

, and

dh/dt=1

. Find

dV/dt

$\frac{dV}{dt}=15\pi\ \text{units}^3/\text{min}$

5For

x^3+y^3=6xy

, find

dy/dx

and then the slope at

(3,3)

$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x},\qquad m_{(3,3)}=-1$

07 · Revisit

Transcript and assigned practice

Search the corrected transcript for a rule or worked example. Captions follow the edited video timeline.

Download corrected transcript

Return to Friday 9/25 in eKadence for Stewart Section 2.5 problems 2, 3, 7, 9, 13, 15, 25, and 32.