Implicit Differentiation

By the end of this lesson, you will be able to differentiate implicitly defined functions and use the result to find slopes of tangent lines to curves that are not functions.

What Is Implicit Differentiation?

A function like $y = x^2 + 3$ is explicit — y is isolated on the left.

A relation like $x^2 + y^2 = 25$ (a circle) is implicit — x and y are mixed together. You cannot easily solve for y (the top half is $y = \sqrt{25-x^2}$ and the bottom half is $y = -\sqrt{25-x^2}$).

Implicit differentiation lets you find $\frac{dy}{dx}$ without ever solving for y.

The Technique

Think of y as a function of x (even if we haven't solved for it). Then differentiate both sides with respect to x, applying the chain rule whenever y appears.

Key: $\frac{d}{dx}[y] = \frac{dy}{dx}$ and $\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}$ (chain rule — y is the inside function).

After differentiating, solve algebraically for $\frac{dy}{dx}$.

Example 1: Circle

Differentiate $x^2 + y^2 = 25$ implicitly.

$\frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = \frac{d}{dx}[25]$

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

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

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

Interpretation: At the point $(3, 4)$ (on the circle since $9+16=25$):

$\frac{dy}{dx} = -\frac{3}{4}$

The tangent line at $(3, 4)$ has slope $-3/4$.

Example 2: More Complex

Differentiate $x^3 + y^3 = 6xy$ (the folium of Descartes).

$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$

Collect $\frac{dy}{dx}$ terms:

$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$

$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$

$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$

Example 3: With Trig

Differentiate $\sin(xy) = x$.

Left side: Apply chain rule — the argument of sin is $xy$, a product.

$\frac{d}{dx}[\sin(xy)] = \cos(xy) \cdot \frac{d}{dx}[xy] = \cos(xy)\left(y + x\frac{dy}{dx}\right)$

So: $\cos(xy)\left(y + x\frac{dy}{dx}\right) = 1$

$y\cos(xy) + x\cos(xy)\frac{dy}{dx} = 1$

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