Implicit differentiation is a different way to find the derivative when $y$ is “stuck” inside an equation and you can’t (or don’t want to) isolate it as $y = \dots$ .

Always treat $y$ as a function of $x$ .

Every time you differentiate a term containing $y$ , you must multiply by $\frac{dy}{dx}$ because of the Chain Rule.

Implicit Differentiation - Textbook Content 1.png

Strategy

Differentiate both sides of the equation with respect to $x$ .
Whenever you differentiate $x$ , do it normally.
Whenever you differentiate $y$ , do it normally but attach a $\frac{dy}{dx}$ to it.
Rearrange the equation to solve for $\frac{dy}{dx}$ .

Example

$x = \frac{2y}{x^2 - y}$

x (x^{2} - y) = 2 y x^{3} - x y = 2 y x^{3} - x y - 2 y = 0 (applied product rule) 3 x^{2} - (x \frac{d y}{d x} + y) - 2 \frac{d y}{d x} = 0 - x \frac{d y}{d x} - y - 2 \frac{d y}{d x} = - 3 x^{2} - x \frac{d y}{d x} - 2 \frac{d y}{d x} = - 3 x^{2} + y \frac{d y}{d x} = \frac{d y}{d x} = \frac{- 3 x ^{2} + y}{- x - y}

Product rule working out:

u = x v = y \frac{d u}{d x} = x^{0} = 1 \frac{d v}{d y} = y^{0} = 1 \times \frac{d y}{d x} = u \frac{d v}{d y} + v \frac{d u}{d x} = x \frac{d y}{d x} + y

Implicit Differentiation

Strategy

Example

x=2yx2−yx = \frac{2y}{x^2 - y}x=x2−y2y​

$x = \frac{2y}{x^2 - y}$