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.
• Calculate higher-order derivatives implicitly.

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:

$$x{y}^2+y=1$$

This equation describes a curve, not a function, because some $x$-values correspond to more than one $y$-value. If you graph it, it fails the vertical line test.

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

Step-by-step process

1. Start with the equation:

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 $x$.

For example, a circle of radius 5 can’t be described by a single function $y=f(x)$, but it can be described by the equation:

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

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 $y$, multiply by $y^{\prime }$ (or $\frac{dy}{dx}$) right away.

For the circle:

$$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(25)$$

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

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

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

Why does this work?

Implicit differentiation is really 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, we usually drop the explicit $(x)$ and write:

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

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.

On the AP exam, you’ll often use 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

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

Example

Find where the curve

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

has horizontal or vertical tangent lines.

Solution

Answer:

• Horizontal: $(\frac{1}{3},-\frac{1}{3})$ and $(1,1)$
• Vertical: $(0,0)$ and $(\frac{4}{3},\frac{2}{3})$

Differentiate implicitly:

$$2x+2y\cdot \frac{dy}{dx}=x\cdot \frac{dy}{dx}+y+1$$

Now solve for $\frac{dy}{dx}$:

$$2y\cdot \frac{dy}{dx}-x\cdot \frac{dy}{dx}=-2x+y+1$$

$$\frac{dy}{dx}(2y-x)=y+1-2x$$

$$\frac{dy}{dx}=\frac{-2x+y+1}{2y-x}$$

1. Horizontal tangent line(s):

Horizontal tangents occur when the numerator is $0$ (and the denominator is not $0$).

Set the numerator equal to $0$:

$$-2x+y+1=0$$

$$y=2x-1$$

and require the denominator

$$2y-x\ne 0$$

To find the points on the curve where $y=2x-1$, substitute into the original equation and solve for $x$:

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

$$x^2+(2x-1{)}^2=x(2x-1)+x$$

$$x^2+(2x-1{)}^2=2x^2-x+x$$

$$x^2+(2x-1{)}^2-2x^2=0$$

$$(2x-1{)}^2-x^2=0$$

Factor as a difference of squares:

$$[(2x-1)+x][(2x-1)-x]=0$$

$$(3x-1)(x-1)=0$$

$$x=\frac{1}{3},1$$

Now find the corresponding $y$-values using $y=2x-1$:

$y=-\frac{1}{3}$ for $x=\frac{1}{3}$ and

$y=1$ for $x=1$

Finally, check the denominator $2y-x$ at each point. Plugging in $(\frac{1}{3},-\frac{1}{3})$ and $(1,1)$ gives nonzero values, so both points are valid. Therefore, the tangent line is horizontal at both points.

2. Vertical tangent line(s):

Vertical tangents occur when the denominator is $0$ (and the numerator is not $0$).

Set the denominator equal to $0$:

$$\begin{gathered} 2y-x=0 \\ x=2y \end{gathered}$$

and require the numerator

$$-2x+y+1\ne 0$$

Substitute $x=2y$ into the original equation:

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

$$(2y{)}^2+y^2=(2y)y+2y$$

$$4y^2+y^2=2y^2+2y$$

$$5y^2-2y^2-2y=0$$

$$3y^2-2y=0$$

$$y(3y-2)=0$$

$$y=0,\frac{2}{3}$$

Using $x=2y$, the candidate points are $(0,0)$ and $(\frac{4}{3},\frac{2}{3})$.

Check the numerator $-2x+y+1$ at each point. Neither makes the numerator $0$, so both points are valid. Therefore, the tangent line is vertical at both points.

Higher order derivatives

So far, we’ve found first derivatives. A first derivative tells you how $y$ changes as $x$ changes (graphically, whether the curve is increasing or decreasing as you move left to right).

The second derivative is the derivative of the first derivative. It tells you how the rate of change is changing.

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

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

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

A major application of the second derivative comes later. For now, let’s find the second derivative of a curve using implicit differentiation.

Use the circle from earlier:

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

We already found

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

Differentiate both sides again with respect to $x$:

$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(-\frac{x}{y})$$

The left side becomes $\frac{d^{2}y}{d{x}^2}$, which is the notation for the second derivative.

On the right side, use the quotient rule. Remember: when you differentiate a term involving $y$ with respect to $x$, you need the chain rule, so a factor of $\frac{dy}{dx}$ appears.

$$\frac{d^{2}y}{d{x}^2}=-\left(\frac{y\cdot 1-x\cdot 1\cdot \frac{dy}{dx}}{y^2}\right)$$

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

Now substitute $-\frac{x}{y}$ for $\frac{dy}{dx}$:

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

$$=\frac{-y-\frac{x^2}{y}}{y^2}$$

Simplify:

$$\frac{d^{2}y}{d{x}^2}=\frac{-\frac{y^2}{y}-\frac{x^2}{y}}{y^2}$$

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

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

Because the original equation tells us $x^2+y^2=25$, we can also write:

$$\frac{d^{2}y}{d{x}^2}=-\frac{25}{y^3}$$

Later, you’ll use second derivatives like this to analyze the curve’s shape (here, the curve is a circle).

Challenge problems

1. Find where the curve

$$\ln (xy)=y$$

has horizontal or vertical tangent lines.

Solution

2. Consider the curve defined by the equation:

$$a{x}^2+bxy+y^2=12$$

where $a$ and $b$ are constants. If the tangent line to the curve at the point $(2,4)$ has a slope of $1,$ find the values of $a$ and $b$.

Solution