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.
• Calculate higher-order derivatives implicitly.

So far, we’ve worked with functions that explicitly define $y$ in terms of $x$, which means every input $x$ produces only one output value $y$.

However, not all equations define $y$ explicitly. Consider the equation:

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

This is a curve and not a function because some inputs $x$ correspond to multiple output values of $y$. Graphing it shows that it does not pass the vertical line test.

Finding the derivative of equations like these requires a process called implicit differentiation.

Step-by-step process

1. Start with the equation:

To recognize the types of 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

Then the equation is probably a curve that fails the vertical line test and is not a function. For example, a circle of radius 5 can’t be described by a single function, but by the equation:

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

2. Differentiate both sides:

Use basic derivative rules on both $y$ and $x$ but whenever you take the derivative of any term with $y$ in it, immediately multiply it by $y^{\prime }$ (or $\frac{dy}{dx}$).

Differentiating the equation of the circle above:

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

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

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 relies on the chain rule, which applies to composite functions.

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

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

Similarly, being asked to differentiate $y^2$ with respect to $x$ implies that output $y$ changes as $x$ changes. So $y$ is treated as a function of $x$.

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

But the inner “$x$” can be dropped as a shorthand notation, and $y^{\prime }(x)$ is also $\frac{dy}{dx}$, so

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

AP tip:

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

On the AP exam, expect to apply 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 $(\frac{4}{3},\frac{2}{3})$

Differentiating implicitly,

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

Solving 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):

These occur when the numerator equals $0$

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

$$y=2x-1$$

and the denominator

$$2y-x\ne 0$$

To find the points that satisfy $y=2x-1$, substitute back 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 this difference of squares as a shortcut:

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

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

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

Then the corresponding $y$-coordinates, using $y=2x-1$, are:

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

$y=1$ for $x=1$

Plugging the points $(\frac{1}{3},-\frac{1}{3})$ and $(1,1)$ individually into the denominator of the derivative $2y-x$ results in non-zero values. So both of these points are valid and the line tangent to each is horizontal.

2. Vertical tangent line(s):

These occur when the numerator

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

and the denominator

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

Plugging $x=2y$ back 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 potential points are $(0,0)$ and $(\frac{4}{3},\frac{2}{3})$. Checking the denominator $-2x+y+1$, neither of those points result in $0$ so both are valid and the tangent line to the curve at each is vertical.

Higher order derivatives

So far, we’ve only calculated 1st derivatives, which give information on how a function is changing (or graphically, whether it’s increasing or decreasing as you move from left to right). The 2nd derivative, found by differentiating the 1st derivative, tells us how the rate of change itself is changing. For example, if $f(x)=x^2$, then

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

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

A very important application of the 2nd derivative will be covered in later sections, but for now let’s find the 2nd derivative of a curve that requires implicit differentiation.

Take the equation of the circle of radius 5 from the beginning:

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

Its derivative was shown to be

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

The 2nd derivative is found by taking the derivative of 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 2nd derivative. Although the two fractions aren’t actually being multiplied, the fact that it appears that way makes it easier to remember.

On the right side, apply the quotient rule. Remember that differentiating any term with $y$ with respect to $x$ requires the chain rule (so multiply by $\frac{dy}{dx}$ immediately after).

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

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

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

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

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

Simplifying,

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

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

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

Lastly, we could substitute $25$ for $x^2+y^2$ since that was defined by the original equation, so that

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

Later on, we’ll explore how this 2nd derivative gives insight into the curve’s shape (which as we know is a circle here).

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