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## Calculus 1

### Course: Calculus 1>Unit 3

Lesson 3: Implicit differentiation

# Implicit differentiation review

Review your implicit differentiation skills and use them to solve problems.

## How do I perform implicit differentiation?

In implicit differentiation, we differentiate each side of an equation with two variables (usually x and y) by treating one of the variables as a function of the other. This calls for using the chain rule.
Let's differentiate x, squared, plus, y, squared, equals, 1 for example. Here, we treat y as an implicit function of x.
\begin{aligned} x^2+y^2&=1 \\\\ \dfrac{d}{dx}(x^2+y^2)&=\dfrac{d}{dx}(1) \\\\ \dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(y^2)&=0 \\\\ 2x+2y\cdot\dfrac{dy}{dx}&=0 \\\\ 2y\cdot\dfrac{dy}{dx}&=-2x \\\\ \dfrac{dy}{dx}&=-\dfrac{x}{y} \end{aligned}
Notice that the derivative of y, squared is 2, y, dot, start fraction, d, y, divided by, d, x, end fraction and not simply 2, y. This is because we treat y as a function of x.
Want a deeper explanation of implicit differentiation? Check out this video.