# Implicit Differentiation

So far we have done calculus with explicit functions. That is we have a function $f$ explicitly in terms of x. For example, $f(x) = \sin x + 3x$ is a function explicitly in terms of $x$. However, sometimes we have a curve given by an equation that is not explicitly in terms of just $x$. For example $x^2 + y^2 = 1$, or $x^y - y + x = 3y^x$. We can use a technique known as **implicit differentiation** to differentiate these curves.

Recall that if we have an explicit function, for example $y = x^3$, then it follows that:

(1)What we are really doing is differentiating both sides of the equation. The derivative of $y$ is $\frac{dy}{dx}$, while the derivative of $x^3$ is $3x^2$ and:

(2)We already know this of course, so we can use this intuition to differentiate implicit equations. For example, let's differentiate the $y^2 - x = x^2 - 2y$

Let's first differentiate both sides of the equation, that is:

(3)Recall that $y^2 = y \cdot y$, so we can use the Derivative Chain Rule to differentiate this:

(4)Hence we get that $\frac{d}{dx} y^2 = 2y \cdot y'$, and substituting this in and isolating for $y'$ we get:

(5)Note: We differentiated $y^2$ in this example and got $2y \cdot y'$. We can of course generalize this in differentiating $y^n$ for $n \in \mathbb{Z}^+$ (the symbol $\mathbb{Z}^+$ represents the set of positive integers, for example $1, 2, 3, ...$) with the formula $\frac{d}{dx} y^n = ny^{n-1} \cdot y'$. |

We will now look at some examples of implicit differentiation.

## Example 1

**Differentiate $xy - x^3 = y^3 - \cos y$.**

We will first differentiate both sides of the equation to get:

(6)We will need to use the chain rule to differentiate $xy$ and $\cos y$:

(7)Then let's isolate $y'$:

(8)## Example 2

**Differentiate the $x^2 + \cos y = y$.**

Taking the derivative of both sides and applying the chain rule we obtain:

(9)