Implicit Differentiation

106. Implicit Differentiation

When a function is not explicitly solved for one variable in terms of another. E.g.:

𝑥^{2} + 𝑦^{2} = 1

Instead of solving for $𝑦$ explicitly in terms of $𝑥$ , implicit differentiation allows you to differentiate both sides of an equation directly, treating $𝑦$ as an implicit function of $𝑥$ .

Steps for Implicit Differentiation

Differentiate both sides of the equation with respect to $𝑥$ , treating y as a function of $𝑥$
Apply the chain rule whenever differentiating $𝑦$ , since $𝑦 = 𝑦 (𝑥)$
Solve for $\frac{𝑑 𝑥}{𝑑 𝑦}$

Example

𝑥^{2} + 𝑦^{2} = 1

Differentiate both sides

\begin{aligned} \frac{𝑑}{𝑑 𝑥} [𝑥^{2} + 𝑦^{2}] & = \frac{𝑑}{𝑑 𝑥} [1] \\ \frac{𝑑}{𝑑 𝑥} [𝑥^{2}] + \frac{𝑑}{𝑑 𝑥} [𝑦^{2}] & = \frac{𝑑}{𝑑 𝑥} [1] \\ \frac{𝑑}{𝑑 𝑥} [𝑥^{2}] + \frac{𝑑}{𝑑 𝑥} [𝑦^{2}] & = 0 \end{aligned}

Apply the chain rule to $𝑦^{2}$

2 𝑥 + 2 𝑦 \frac{𝑑 𝑦}{𝑑 𝑥} = 0

Slove for $\frac{𝑑 𝑦}{𝑑 𝑥}$

\begin{aligned} 2 𝑦 \cdot \frac{𝑑 𝑦}{𝑑 𝑥} & = - 2 𝑥 \\ \frac{𝑑 𝑦}{𝑑 𝑥} & = \frac{- 2 𝑥}{2 𝑦} \\ \frac{𝑑 𝑦}{𝑑 𝑥} & = - \frac{𝑥}{𝑦} \end{aligned}

\begin{matrix} 𝑥^{2} + 𝑦^{2} = 5^{2} \\ 2 𝑥 𝑑 𝑥 + 2 𝑦 𝑑 𝑦 = 0 \\ \frac{𝑑 𝑦}{𝑑 𝑥} = - \frac{𝑥}{𝑦} \end{matrix}