Implicit differentiation

What you’ll learn

• How to find $\frac{dy}{dx}$ with implicit differentiation when $x$ and $y$ are mixed.
• Recognize when an equation defines $y$ implicitly (not as a function).
• Identify horizontal and vertical tangent lines using the derivative.

So far, we’ve worked with functions that explicitly define $y$ in terms of $x$. In other words, each input $x$ produces exactly one output value $y$.

However, not all equations define $y$ explicitly. Consider the equation for a circle of radius $5$:

$$x^2+y^2=25$$

This equation does not define $y$ as a function of $x$, because some $x$-values correspond to more than one $y$-value (the graph fails the vertical line test).

To find derivatives for equations like this, we use implicit differentiation.

How to find $\frac{dy}{dx}$

Step 1: Recognizing implicit equations

To recognize equations that require implicit differentiation, look for:

• Terms with $x$ and $y$ multiplied together
• $y$ appearing multiple times in the equation
• $y$ being raised to a power somewhere

When you see these features, the equation often represents a curve that fails the vertical line test, so it isn’t a function of $x$.

Step 2: Differentiate both sides

Differentiate with respect to $x$ on both sides. Use the usual derivative rules, but every time you differentiate a term involving $y$, immediately multiply it by $y^{\prime }$ (or $\frac{dy}{dx}$).

For example, to implicitly differentiate the equation of a circle:

$$\begin{array}{rl}{x}^2+y^2& =25\\ \frac{d}{dx}(x^2+y^2)& =\frac{d}{dx}(25)\\ 2x+2y\cdot \frac{dy}{dx}& =0\end{array}$$

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$:

$$\begin{array}{rl}2y\cdot \frac{dy}{dx}& =-2x\\ \frac{dy}{dx}& =-\frac{2x}{2y}\\ & =\boxed{-\frac{x}{y}}\end{array}$$

Why does this work?

Implicit differentiation is the chain rule in disguise.

For example, if $f(g(x))=(g(x){)}^2$, then

$$f^{\prime }(x)=2(g(x))\cdot g^{\prime }(x)$$

When you differentiate $y^2$ with respect to $x$, you’re treating $y$ as a function of $x$ (even if it isn’t written as $y(x)$). So:

$$\frac{d}{dx}(y(x){)}^2=2(y(x))\cdot y^{\prime }(x)$$

In practice, the inner $x$ is not written, and

$$\frac{d}{dx}(y^2)=2y\cdot \frac{dy}{dx}$$

AP tip:

Don’t forget the chain rule. If the problem asks for the derivative of $y$ with respect to $x$, then every time you differentiate a term involving $y$, multiply by $y^{\prime }$ or $\frac{dy}{dx}$ immediately.

Slope of a tangent line

If asked to find the slope of the tangent line at a particular point, it often helps to substitute the values immediately after differentiating, before rearranging to solve for $\frac{dy}{dx}$.

Find the equation of the line tangent to the curve $y^2-xy=2x$ at the point where $y=2$.

Solution

Horizontal and vertical tangent lines

Many past AP exams contained problems that required implicit differentiation to identify tangent lines to curves.

1. Horizontal tangent lines occur when $\frac{dy}{dx}=0$. More specifically,

$$\frac{dy}{dx}=\frac{0}{\ne 0}$$

2. Vertical tangent lines occur when $\frac{dy}{dx}$ is undefined. More specifically,

$$\frac{dy}{dx}=\frac{\ne 0}{0}$$

3. No tangent line exists where

$$\frac{dy}{dx}=\frac{0}{0}$$

This case (numerator and denominator both zero) requires further analysis - the point may be a cusp or a node on the curve.

Example

Find where the curve

$$x^2+y^2=xy+x$$

has horizontal or vertical tangent lines.

Answers

Solutions

a) Horizontal tangent lines

b) Vertical tangent lines