Introduction

Up to this point, we've differentiated functions by isolating for $y$ and then taking the derivative of the other side. But it's not always easy — or even possible — to solve directly for $y$. This is where implicit differentiation comes in.

The idea is to differentiate both sides of an equation with respect to $x$ and apply the chain rule as needed. This works because differentiation is an operation just like addition or subtraction: if you differentiate both sides of a true equation, it remains true.

The key is that whenever you differentiate a term involving $y$, you must multiply by $\dfrac{dy}{dx}$ (also written $y'$) via the chain rule, since $y$ is itself a function of $x$.

Examples

Example 1: Find $\dfrac{dy}{dx}$ given $x^2 + y^2 = 1$.

Solution: Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(1)$

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

Solving for $\dfrac{dy}{dx}$:

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

In the second step, we used the chain rule on $y^2$: since $y$ is a function of $x$, differentiating gives $2y \cdot \dfrac{dy}{dx}$.

Example 2: Find $y'$ given $x^2 + xy + y^3 = 0$.

Solution: The middle term $xy$ is a product of two functions of $x$, so we need both the product rule and the chain rule. Differentiating both sides:

$\frac{d}{dx}(x^2 + xy + y^3) = \frac{d}{dx}(0)$

$2x + (x \cdot y' + 1 \cdot y) + 3y^2 y' = 0$

Collecting all $y'$ terms on one side:

$y'(x + 3y^2) = -2x - y$

$\boxed{y' = \frac{-2x - y}{x + 3y^2}}$

Practice Problems

Problem 1: Given $2xy + x^3 - 3y^2 = 5$, find $\dfrac{dy}{dx}$.

Answer: $\dfrac{dy}{dx} = \dfrac{-2y - 3x^2}{2x - 6y}$

Problem 2: Given $x^3 y + y^2 - x^2 = 5$, find the slope of the tangent line at the point $(2, 1)$.

Answer: $-\dfrac{4}{5}$

Problem 3: Given $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}$, find $\dfrac{dy}{dx}$.

Answer: $-\dfrac{y^2}{x^2}$