Implicit differentiation

Warm-up

Three quick questions from earlier lessons.

Discover

The circle $x^2 + y^2 = 25$ is not the graph of one function, but it has a tangent line at each point. If $y$ depends on $x$ , what is $\dfrac{d}{dx}\big(y^2\big)$ ?

$2y \cdot \dfrac{dy}{dx}$
$2y$

Key idea

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Implicit differentiation: every term with

y

picks up a factor

\dfrac{dy}{dx}

(chain rule); products like

xy

need the product rule. Then solve the equation for

\dfrac{dy}{dx}

.

Worked example

$x^2 + y^2 = 25$ . Find $\dfrac{dy}{dx}$ .

2x + 2y\,y' = 0

Differentiate each term; (y²)' = 2y y'.

y' = -\dfrac{x}{y}

Isolate y'.

y'(3, 4) = -\dfrac{3}{4}

At (3, 4): the tangent's slope.

Your turn, guided

Your turn: $y^3 + x = 7$ .

3y^2\,y' + 1 = 0

(y³)' = 3y² y'; (x)' = 1.

y' = -\dfrac{1}{3y^2}

Isolate y'.

Practice ★

$x^2 + 6y^2 = 63$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

$y^2 = x^3 + 5x$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$y^2 = x^3 + 6x$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Watch out!

Practice ★★

$xy + 7x = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$xy + 3x = 15$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Exit ticket

Three questions to check you have got it.

$x^2 + 6y^2 = 63$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 19$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .
$x^2 + 7y^2 = 69$ . Find $\dfrac{dy}{dx}$ in terms of $x$ and $y$ .

Summary card

Implicit differentiation

Differentiate both sides with respect to $x$ .
Each $y$ term gets a factor $y'$ ; use the product rule for $xy$ .
Solve for $y'$ .

Watch out: $(y^2)' = 2y\,y'$ , not $2y$ .

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