Derivatives of inverse functions | Advanced differentiation

What you’ll learn:

How to find the slope of the tangent line to the inverse function at a given point

Suppose $f (x) = x^{5} + x$ . To find its inverse, swap the $x$ and $y$ so that

$x = y^{5} + y$

and solve for $y$ , which would be $f^{- 1} (x)$ .

Graphically, a one-to-one function (passing the horizontal line test) and its inverse are reflections of each other over the line $y = x$ line, which swaps the $x$ and $y$ -values as well as the domain and range. This means that if $f (3) = 5$ , then $f^{- 1} (5) = 3$ .

But the process of actually finding the inverse function in order to apply calculus to it can become tedious the more complicated $f (x)$ is. Instead, there’s a formula that directly evaluates the derivative of a function’s inverse at a particular point.

Given $f (x)$ , the derivative of its inverse $f^{- 1} (x)$ is derived using this fundamental property of inverse functions:

$f (f^{- 1} (x)) = x$

Differentiating both sides with respect to $x$ ,

$\frac{d}{d x} (f (f^{- 1} (x)) = \frac{d}{d x} (x)$

Expanding using the chain rule,

$f^{'} (f^{- 1} (x)) \cdot \frac{d}{d x} (f^{- 1} (x)) = 1$

After dividing by $f^{'} (f^{- 1} (x)$ ,

$\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))}$

It’s much easier to remember this formula if you let $g (x)$ be $f^{- 1} (x)$ , the inverse function of $f (x)$ . Then rewritten,

$\frac{d}{d x} (g (x)) = \frac{1}{f ^{'} ( g ( x ))}$

Derivative of an inverse function:

If $g (x)$ is the inverse of $f (x)$ , then

$g^{'} (x) = \frac{1}{f ^{'} ( g ( x ))}$

Step-by-step process

A typical problem on the AP exam will define a function $f (x)$ and request the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , or $g^{'} (a)$ . For example,

Let $f (x) = x^{3}$ and $g (x)$ be the inverse of $f (x)$ . Find the slope of the tangent line to $g (x)$ at $x = 8$ .

Follow these steps to find $g^{'} (a)$ , or $g^{'} (8)$ , in this example problem:

1. Write the formula for clarity:

For the derivative of the inverse function at $x = 8$ :

$g^{'} (8) = \frac{1}{f ^{'} ( g ( 8 ))}$

2. Find $g (8)$ :

Start with the innermost function in the denominator on the right. Since $f (x)$ and $g (x)$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set the defined function $f (b)$ equal to $8$ and solve for $b$ .

$b^{3} = 8 b = 2$

$f (2) = 8 ⟺ g (8) = 2$

Plugging $g (8) = 2$ into the formula, we have, so far:

$g^{'} (8) = \frac{1}{f ^{'} ( 2 )}$

3. Find $f^{'} (x)$ and evaluate:

The derivative of $f (x) = x^{3}$ is $f^{'} (x) = 3 x^{2}$ .

Then

$f^{'} (2) = 3 (2)^{2} = 12$

4. Compute $g^{'} (a)$ :

$g^{'} (a)$ is the reciprocal of $f^{'} (b)$ .

$g^{'} (8) = \frac{1}{12}$

To summarize, if $f (x)$ and $g (x)$ are inverses with point $(a, b)$ on $g (x)$ , then to find $g^{'} (a)$ :

Write the inverse derivative formula for clarity.
Find $g (a)$
Find $f^{'} (b)$
Compute $g^{'} (a)$

Sidenote

On notation

Watch out for notation. These two are different:

1. $f^{- 1} (x)$

Means inverse of $f (x)$

2. $(f (x))^{- 1}$

Means reciprocal of $f (x) = \frac{1}{f ( x )}$

Examples

1. Let $f (x)$ and $g (x)$ be differentiable functions such that $f (g (x)) = x$ .

Find the equation of the tangent line at $x = 2$ on $g (x)$ if

$f (x) = x^{3} - x^{2} + 2 x$

Solution

(spoiler)

Sometimes a problem on the AP exam will not state that the two are inverses. Instead, the key phrase to recognize is $f (g (x)) = x$ that signals the use of inverse functions.

1. Write the formula:

To find the equation of the tangent line to $g (x)$ at $x = 2$ , we need $g^{'} (2)$ :

$g^{'} (2) = \frac{1}{f ^{'} ( g ( 2 ))}$

2. Find $g (2)$ :

Since $g (x)$ and $f (x)$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (you may use $x$ or $b$ depending on what you find easier to work with).

Solving the equation for $x$ ,

$x^{3} - x^{2} + 2 x = 2$

$x^{3} - x^{2} + 2 x - 2 = 0$

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it means that $g (2) = 1$ , and the next step in the formula becomes

$g^{'} (2) = \frac{1}{f ^{'} ( 1 )}$

3. Find $f^{'} (x)$ and evaluate

Since $f (x) = x^{3} - x^{2} + 2 x$ ,

$f^{'} (x) = 3 x^{2} - 2 x + 2$

Then

$f^{'} (1) = 3 - 2 + 2$

$= 3$

4. Compute $g^{'} (a)$

$g^{'} (2) = \frac{1}{3}$

From step 2, we calculated $g (2) = 1$ , which is the point of tangency for this problem. Then the equation of the tangent line is

$y - 1 = \frac{1}{3} (x - 2)$

To verify this in Desmos, type the equations

1. $y = x^{3} - x^{2} + 2 x$

This is $f (x)$

2. $x = y^{3} - y^{2} + 2 y$

This is $g (x)$ , the inverse

3. Equation of tangent line

Verify it to be the tangent line to $g (x)$ at $(2, 1)$

2. Suppose that $g (x) = f^{- 1} (x)$ , where $f (x)$ is a differentiable function. Given the following:

$f (2) = 5$

$f^{'} (2) = 7$

find $g^{'} (5)$ .

Solution

(spoiler)

Answer: $\frac{1}{7}$

1. Write the formula:

$g^{'} (5) = \frac{1}{f ^{'} ( g ( 5 ))}$

2. Find $g (5)$ :

$f (x)$ and $g (x)$ are inverses and from the given information

$f (2) = 5$

So it follows that

$g (5) = 2$

Then

$g^{'} (5) = \frac{1}{f ^{'} ( 2 )}$

3. Find $f^{'} (x)$ and evaluate

The problem also stated that $f^{'} (2) = 7$ .

4. Compute $g^{'} (a)$

Lastly, $g^{'} (5)$ is the reciprocal of this.

$g^{'} (5) = \frac{1}{7}$

Table problem

You may also encounter problems with a table to extract information from.

Values of differentiable function $f (x)$ and its derivative $f^{'} (x)$ are given below. Let $g (x)$ be the inverse of $f (x)$ .

$x$ $f (x)$ $f^{'} (x)$

$2$ $10$ $5$

$3 $15$ $6$

$4$ $20$ $8$

$5$ $30$ $10$

$6$ $50$ $12$

a) Find $g (15) .$
b) Find $g^{'} (15) .$

$x$	$f (x)$	$f^{'} (x)$
$2$	$10$	$5$
$3	$15$	$6$
$4$	$20$	$8$
$5$	$30$	$10$
$6$	$50$	$12$

Solutions

a) Find $g (15)$ .

Because $g$ and $f$ are inverses,

$g (15) = b ⟺ f (b) = 15$

Locate where $f (x) = 15$ is in the table, which is when $x = 3$ . So

$g (15) = 3$

b) Find $g^{'} (15)$ .

1. Write formula

$g^{'} (15) = \frac{1}{f ^{'} ( g ( 15 ))}$

2. Find $g (15)$ .

From part a, we found $g (15) = 3$ . Then

$g^{'} (15) = \frac{1}{f ^{'} ( 3 )}$

3. Find $f^{'} (3)$ .

From the table, $f^{'} (3) = 6$ .

4. Compute $g^{'} (15)$ .

It’s the reciprocal.

$g^{'} (15) = \frac{1}{6}$

Let’s try a more challenging one.

Values of differentiable function $f (x)$ and its derivative $f^{'} (x)$ are given below, with some entries intentionally left blank. Let $g (x) = f^{- 1} (x)$ .

$x$ $f (x)$ $f^{'} (x)$

$1$ $2$

$3$ $6$ $4$

$5$ $5$

$7$ $18$

$9$ $25$ $10$

a) Find $g^{'} (6)$ .

b) Determine $f^{'} (1)$ , given that $g^{'} (2) = \frac{1}{8}$ .

$x$	$f (x)$	$f^{'} (x)$
$1$	$2$
$3$	$6$	$4$
$5$		$5$
$7$	$18$
$9$	$25$	$10$

Solutions

a) Find $g^{'} (6)$ .

(spoiler)

With the formula,

$g^{'} (6) = \frac{1}{f ^{'} ( g ( 6 ))}$

$g$ and $f$ are inverses, so

$g (6) = b ⟺ f (b) = 6$

In the table, $f (x) = 6$ when $x = 3$ . Then

$g^{'} (6) = \frac{1}{f ^{'} ( 3 )}$

Lastly, $f^{'} (3) = 4$ from the table. So

$g^{'} (6) = \frac{1}{4}$

b) Determine $f^{'} (1)$ given that $g^{'} (2) = \frac{1}{8}$ .

(spoiler)

This problem requires a bit of working backwards.

From the formula,

$g^{'} (2) = \frac{1}{f ^{'} ( g ( 2 ))}$

Which means that $f^{'} (g (2)) = 8$ . Next,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (x) = 2$ when $x = 1$ . So

$f (1) = 2 ⟺ g (2) = 1$

Therefore $f^{'} (g (2)) = f^{'} (1) = 8$ .

Key points

The slope of the tangent line to $f^{- 1} (x)$ at $(a, b)$ is the reciprocal of the slope of the tangent line to $f (x)$ at $(b, a)$ .
$f (g (x)) = x$ (or $g (f (x)) = x$ is a key phrase that signals the topic is about inverse functions.

What you’ll learn:

How to find the slope of the tangent line to the inverse function at a given point

Suppose $f (x) = x^{5} + x$ . To find its inverse, swap the $x$ and $y$ so that

$x = y^{5} + y$

and solve for $y$ , which would be $f^{- 1} (x)$ .

Given $f (x)$ , the derivative of its inverse $f^{- 1} (x)$ is derived using this fundamental property of inverse functions:

$f (f^{- 1} (x)) = x$

Differentiating both sides with respect to $x$ ,

$\frac{d}{d x} (f (f^{- 1} (x)) = \frac{d}{d x} (x)$

Expanding using the chain rule,

$f^{'} (f^{- 1} (x)) \cdot \frac{d}{d x} (f^{- 1} (x)) = 1$

After dividing by $f^{'} (f^{- 1} (x)$ ,

$\frac{d}{d x} (f^{- 1} (x)) = \frac{1}{f ^{'} ( f ^{- 1} ( x ))}$

It’s much easier to remember this formula if you let $g (x)$ be $f^{- 1} (x)$ , the inverse function of $f (x)$ . Then rewritten,

$\frac{d}{d x} (g (x)) = \frac{1}{f ^{'} ( g ( x ))}$

Derivative of an inverse function:

If $g (x)$ is the inverse of $f (x)$ , then

$g^{'} (x) = \frac{1}{f ^{'} ( g ( x ))}$

Step-by-step process

A typical problem on the AP exam will define a function $f (x)$ and request the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , or $g^{'} (a)$ . For example,

Let $f (x) = x^{3}$ and $g (x)$ be the inverse of $f (x)$ . Find the slope of the tangent line to $g (x)$ at $x = 8$ .

Follow these steps to find $g^{'} (a)$ , or $g^{'} (8)$ , in this example problem:

1. Write the formula for clarity:

For the derivative of the inverse function at $x = 8$ :

$g^{'} (8) = \frac{1}{f ^{'} ( g ( 8 ))}$

2. Find $g (8)$ :

Start with the innermost function in the denominator on the right. Since $f (x)$ and $g (x)$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set the defined function $f (b)$ equal to $8$ and solve for $b$ .

$b^{3} = 8 b = 2$

$f (2) = 8 ⟺ g (8) = 2$

Plugging $g (8) = 2$ into the formula, we have, so far:

$g^{'} (8) = \frac{1}{f ^{'} ( 2 )}$

3. Find $f^{'} (x)$ and evaluate:

The derivative of $f (x) = x^{3}$ is $f^{'} (x) = 3 x^{2}$ .

Then

$f^{'} (2) = 3 (2)^{2} = 12$

4. Compute $g^{'} (a)$ :

$g^{'} (a)$ is the reciprocal of $f^{'} (b)$ .

$g^{'} (8) = \frac{1}{12}$

To summarize, if $f (x)$ and $g (x)$ are inverses with point $(a, b)$ on $g (x)$ , then to find $g^{'} (a)$ :

Write the inverse derivative formula for clarity.
Find $g (a)$
Find $f^{'} (b)$
Compute $g^{'} (a)$

Sidenote

On notation

Watch out for notation. These two are different:

1. $f^{- 1} (x)$

Means inverse of $f (x)$

2. $(f (x))^{- 1}$

Means reciprocal of $f (x) = \frac{1}{f ( x )}$

Examples

1. Let $f (x)$ and $g (x)$ be differentiable functions such that $f (g (x)) = x$ .

Find the equation of the tangent line at $x = 2$ on $g (x)$ if

$f (x) = x^{3} - x^{2} + 2 x$

Solution

(spoiler)

Sometimes a problem on the AP exam will not state that the two are inverses. Instead, the key phrase to recognize is $f (g (x)) = x$ that signals the use of inverse functions.

1. Write the formula:

To find the equation of the tangent line to $g (x)$ at $x = 2$ , we need $g^{'} (2)$ :

$g^{'} (2) = \frac{1}{f ^{'} ( g ( 2 ))}$

2. Find $g (2)$ :

Since $g (x)$ and $f (x)$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (you may use $x$ or $b$ depending on what you find easier to work with).

Solving the equation for $x$ ,

$x^{3} - x^{2} + 2 x = 2$

$x^{3} - x^{2} + 2 x - 2 = 0$

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it means that $g (2) = 1$ , and the next step in the formula becomes

$g^{'} (2) = \frac{1}{f ^{'} ( 1 )}$

3. Find $f^{'} (x)$ and evaluate

Since $f (x) = x^{3} - x^{2} + 2 x$ ,

$f^{'} (x) = 3 x^{2} - 2 x + 2$

Then

$f^{'} (1) = 3 - 2 + 2$

$= 3$

4. Compute $g^{'} (a)$

$g^{'} (2) = \frac{1}{3}$

From step 2, we calculated $g (2) = 1$ , which is the point of tangency for this problem. Then the equation of the tangent line is

$y - 1 = \frac{1}{3} (x - 2)$

To verify this in Desmos, type the equations

1. $y = x^{3} - x^{2} + 2 x$

This is $f (x)$

2. $x = y^{3} - y^{2} + 2 y$

This is $g (x)$ , the inverse

3. Equation of tangent line

Verify it to be the tangent line to $g (x)$ at $(2, 1)$

2. Suppose that $g (x) = f^{- 1} (x)$ , where $f (x)$ is a differentiable function. Given the following:

$f (2) = 5$

$f^{'} (2) = 7$

find $g^{'} (5)$ .

Solution

(spoiler)

Answer: $\frac{1}{7}$

1. Write the formula:

$g^{'} (5) = \frac{1}{f ^{'} ( g ( 5 ))}$

2. Find $g (5)$ :

$f (x)$ and $g (x)$ are inverses and from the given information

$f (2) = 5$

So it follows that

$g (5) = 2$

Then

$g^{'} (5) = \frac{1}{f ^{'} ( 2 )}$

3. Find $f^{'} (x)$ and evaluate

The problem also stated that $f^{'} (2) = 7$ .

4. Compute $g^{'} (a)$

Lastly, $g^{'} (5)$ is the reciprocal of this.

$g^{'} (5) = \frac{1}{7}$

Table problem

You may also encounter problems with a table to extract information from.

Values of differentiable function $f (x)$ and its derivative $f^{'} (x)$ are given below. Let $g (x)$ be the inverse of $f (x)$ .

$x$ $f (x)$ $f^{'} (x)$

$2$ $10$ $5$

$3 $15$ $6$

$4$ $20$ $8$

$5$ $30$ $10$

$6$ $50$ $12$

a) Find $g (15) .$
b) Find $g^{'} (15) .$

$x$	$f (x)$	$f^{'} (x)$
$2$	$10$	$5$
$3	$15$	$6$
$4$	$20$	$8$
$5$	$30$	$10$
$6$	$50$	$12$

Solutions

a) Find $g (15)$ .

Because $g$ and $f$ are inverses,

$g (15) = b ⟺ f (b) = 15$

Locate where $f (x) = 15$ is in the table, which is when $x = 3$ . So

$g (15) = 3$

b) Find $g^{'} (15)$ .

1. Write formula

$g^{'} (15) = \frac{1}{f ^{'} ( g ( 15 ))}$

2. Find $g (15)$ .

From part a, we found $g (15) = 3$ . Then

$g^{'} (15) = \frac{1}{f ^{'} ( 3 )}$

3. Find $f^{'} (3)$ .

From the table, $f^{'} (3) = 6$ .

4. Compute $g^{'} (15)$ .

It’s the reciprocal.

$g^{'} (15) = \frac{1}{6}$

Let’s try a more challenging one.

Values of differentiable function $f (x)$ and its derivative $f^{'} (x)$ are given below, with some entries intentionally left blank. Let $g (x) = f^{- 1} (x)$ .

$x$ $f (x)$ $f^{'} (x)$

$1$ $2$

$3$ $6$ $4$

$5$ $5$

$7$ $18$

$9$ $25$ $10$

a) Find $g^{'} (6)$ .

b) Determine $f^{'} (1)$ , given that $g^{'} (2) = \frac{1}{8}$ .

$x$	$f (x)$	$f^{'} (x)$
$1$	$2$
$3$	$6$	$4$
$5$		$5$
$7$	$18$
$9$	$25$	$10$

Solutions

a) Find $g^{'} (6)$ .

(spoiler)

With the formula,

$g^{'} (6) = \frac{1}{f ^{'} ( g ( 6 ))}$

$g$ and $f$ are inverses, so

$g (6) = b ⟺ f (b) = 6$

In the table, $f (x) = 6$ when $x = 3$ . Then

$g^{'} (6) = \frac{1}{f ^{'} ( 3 )}$

Lastly, $f^{'} (3) = 4$ from the table. So

$g^{'} (6) = \frac{1}{4}$

b) Determine $f^{'} (1)$ given that $g^{'} (2) = \frac{1}{8}$ .

(spoiler)

This problem requires a bit of working backwards.

From the formula,

$g^{'} (2) = \frac{1}{f ^{'} ( g ( 2 ))}$

Which means that $f^{'} (g (2)) = 8$ . Next,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (x) = 2$ when $x = 1$ . So

$f (1) = 2 ⟺ g (2) = 1$

Therefore $f^{'} (g (2)) = f^{'} (1) = 8$ .

Key points

The slope of the tangent line to $f^{- 1} (x)$ at $(a, b)$ is the reciprocal of the slope of the tangent line to $f (x)$ at $(b, a)$ .
$f (g (x)) = x$ (or $g (f (x)) = x$ is a key phrase that signals the topic is about inverse functions.