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Grade 12 Calculus · Lesson 2.590 min

Implicit Differentiation · Follow every y

Differentiate the relation without solving for $y$ first. The Chain Rule from Lesson 2.4 is active whenever $y=y(x)$ , so every differentiated y-function must carry its derivative.

Use Mr. Azzam’s routine to differentiate implicit relations, collect dy/dx terms, and find slopes at points.

See every y-termChain Rule on yCollect + factorProduct Rule with xyTrig + implicitSlope at a point

Key takeaway

The three moves

1 · See it

Identify every term containing $y$ .

2 · Differentiate it

Use the correct rule and attach $\frac{dy}{dx}$ to differentiated y-functions.

3 · Solve it

Collect, factor, isolate, then substitute a point if needed.

Visualize it · the equation hides y, not the slope

Before doing algebra, connect the derivative to geometry. A relation can describe a curve even when $y$ is not isolated.

Interactive lab

Visualize the hidden slope

Move P on the upper half of

x^2+y^2=16

. Predict the tangent direction, including the endpoints.

Point x-coordinate

x = 2.4 · y = 3.20

What is the tangent direction at this point?

Why implicit differentiation works

2x+2y\frac{dy}{dx}=0\quad\Rightarrow\quad\frac{dy}{dx}=-\frac{x}{y}\quad(y\ne0)

Explicit or implicit?

In an explicit equation, $y$ is already isolated. In an implicit equation, $x$ and $y$ can appear together. We can differentiate the relation directly.

Explanation

Explicit

y=x^2+3x

Differentiate the right side normally.

Explanation

Implicit

x^2+y^2=25

Differentiate both sides with respect to $x$ .

Key takeaway

First decision

Do not waste time solving for $y$ unless the problem specifically asks you to. Implicit differentiation is designed to work with the relation as written.

The key idea: y depends on x

Explanation

We differentiate with respect to x. Therefore a function of $y$ is a composition, and the Chain Rule is active.

\frac{d}{dx}(y^n)=ny^{n-1}\frac{dy}{dx}

Explanation

\frac{d}{dx}(y)=\frac{dy}{dx}

Explanation

\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}

Explanation

\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}

Your turn

When differentiating

y^4

with respect to

x

, what must appear?

4y^3

4y^3\frac{dy}{dx}

4x^3

Common mistake. Never write

\frac{d}{dx}(y^n)=ny^{n-1}

. The missing

\frac{dy}{dx}

loses the Chain Rule.

Mr. Azzam’s implicit differentiation routine

STEP 1

Differentiate both sides

Work with respect to

x

STEP 2

Track every y-term

Attach

\frac{dy}{dx}

whenever a function of

y

is differentiated.

STEP 3

Collect derivative terms

Move all terms containing

\frac{dy}{dx}

to one side.

STEP 4

Factor

Take

\frac{dy}{dx}

as a common factor.

STEP 5

Isolate

Divide to solve for

\frac{dy}{dx}

STEP 6

Then substitute

If a point is given, substitute only after the derivative formula is ready.

Your turn

After differentiating, how should you proceed to obtain a general formula for

dy/dx

rather than a slope at one specified point?

Substitute a point immediatelyCancel every

y

Collect all terms containing

\frac{dy}{dx}

on one side

Try it · basic implicit differentiation

Make each decision yourself. The next step appears only after a correct choice.

Worked example

Interactive Example · Circle

x^2+y^2=36

Step 1Step 2Step 3

Differentiate both sides. Which line is correct?

Reference example · several y-terms

y^3+y^2-5y-x^2=-4

3y^2\frac{dy}{dx}+2y\frac{dy}{dx}-5\frac{dy}{dx}-2x=0

\left(3y^2+2y-5\right)\frac{dy}{dx}=2x

\frac{dy}{dx}=\frac{2x}{3y^2+2y-5}

Key takeaway

Pattern to notice

The calculus creates several $\frac{dy}{dx}$ terms. The algebra finishes the job: collect → factor → divide.

When x and y are multiplied

A term such as $xy$ needs the Product Rule because both factors vary with $x$ .

\frac{d}{dx}(xy)=y+x\frac{dy}{dx}

Worked example

Interactive Example · Product Rule inside implicit differentiation

x^2+xy+y^2=12

Step 1Step 2Step 3

What is

\frac{d}{dx}(xy)

Product Rule + implicit differentiation

x^2+xy+y^2=7

2x+\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0

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

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

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

Your turn

Differentiate

xy

with respect to

x

x\frac{dy}{dx}

y+x\frac{dy}{dx}

1\cdot\frac{dy}{dx}

Common mistake. For

xy

, do not write only

x\frac{dy}{dx}

. Product Rule gives

y+x\frac{dy}{dx}

Trig and mixed rules

Worked example

Interactive Example · Trig + Chain Rule

2x+\sin y=4

Step 1Step 2Step 3

Differentiate

\sin y

with respect to

x

A y-term inside a trig function

x+\sin y=2

1+\cos y\frac{dy}{dx}=0

\cos y\frac{dy}{dx}=-1

\frac{dy}{dx}=-\frac{1}{\cos y}=-\sec y

Challenge · Chain Rule outside, Product Rule inside

\cos(xy)=x

-\sin(xy)\left(y+x\frac{dy}{dx}\right)=1

-y\sin(xy)-x\sin(xy)\frac{dy}{dx}=1

\frac{dy}{dx}=\frac{-1-y\sin(xy)}{x\sin(xy)}

Common mistake. Implicit differentiation is not a replacement for previous rules. Power, Product, Trigonometric, and Chain Rules still apply; the extra task is tracking

y=y(x)

Your turn

Differentiate

\sin y

with respect to

x

, where

y=y(x)

\cos y\,\frac{dy}{dx}

\cos y

-\sin y\,\frac{dy}{dx}

\cos x\,\frac{dy}{dx}

Find the slope at a point

Differentiate first · substitute second

x^2+y^2=25

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

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

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

x² + y² = 25 · tangent at (3, 4) · slope −3/4

Explanation

Horizontal tangent

After isolating the derivative as a fraction, look for where the numerator is zero while the denominator is nonzero.

Explanation

Vertical tangent

Look for where the denominator is zero while the numerator is nonzero.

Key takeaway

Exam strategy

Do not substitute the point too early. Find a clean formula for $\frac{dy}{dx}$ first, then evaluate it.

Your turn

Question 1 of 2

For

x^2+y^2=25

, what is the tangent slope at

(3,4)

-\frac43

\frac34

-\frac34

Undefined

Interactive · build the derivative yourself

Interactive lab

Implicit Differentiation Lab

Differentiate, collect, isolate, then evaluate. Each step needs your answer before you continue.

x^2+y^2=25

Step 1 of 5

When isolating y′, the denominator must be nonzero.

What happens when differentiating a function of y with respect to x?

Before you finish · exam checklist

Did I differentiate both sides with respect to x?

Did every differentiated y-function receive its chain factor?

Did I use Product Rule when x and y are multiplied?

Did I collect every dy/dx term before factoring?

Did I isolate dy/dx before substituting the point?

Did I simplify only after the derivative structure was correct?

Find your next lesson

Implicit Differentiation · Follow every y

Visualize it · the equation hides y, not the slope

Visualize the hidden slope

Explicit or implicit?

The key idea: y depends on x

Your turn

Your turn · student questions

Mr. Azzam’s implicit differentiation routine

Your turn

Your turn · student questions

Try it · basic implicit differentiation

Step 1

Step 2

Step 3

When x and y are multiplied

Step 1

Step 2

Step 3

Your turn

Your turn · student questions

Trig and mixed rules

Step 1

Step 2

Step 3

Your turn

Your turn · student questions

Find the slope at a point

Your turn

Your turn · student questions

Interactive · build the derivative yourself

Implicit Differentiation Lab

Before you finish · exam checklist

Keep your unfinished answers?