3.4 Derivatives of inverse functions

Achievable AP Calculus AB

3. Advanced differentiation

Our AP Calculus AB course is currently in development and is a work-in-progress.

Derivatives of inverse functions

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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, you would swap $x$ and $y$ :

$x = y^{5} + y$

Then you’d solve for $y$ , which would give $f^{- 1} (x)$ .

Graphically, a one-to-one function (one that passes the horizontal line test) and its inverse are reflections of each other across the line $y = x$ . Reflecting across $y = x$ swaps the $x$ - and $y$ -values, and it also swaps the domain and range. For example, if $f (3) = 5$ , then $f^{- 1} (5) = 3$ .

In many problems, explicitly solving for the inverse function is more work than you need. Instead, you can use a formula that gives the derivative of the inverse at a specific input value.

Given $f (x)$ , we start from the defining property of inverse functions:

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

Differentiate both sides with respect to $x$ :

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

Now apply the chain rule to the left side:

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

Divide both sides by $f^{'} (f^{- 1} (x))$ :

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

It can be easier to read if you rename the inverse function. Let $g (x) = f^{- 1} (x)$ . Then the same formula becomes

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

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 will define a function $f (x)$ and ask for the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , which is $g^{'} (a)$ . 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 $g^{'} (a)$ (here, $g^{'} (8)$ ):

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 $g (8)$ :

Focus on the inside of the denominator first. Since $f$ and $g$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set $f (b)$ equal to $8$ and solve for $b$ :

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

That means

$f (2) = 8 ⟺ g (8) = 2$

Substitute $g (8) = 2$ into the formula:

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

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

Differentiate $f (x) = x^{3}$ :

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

Evaluate at $x = 2$ :

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

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

Now take the reciprocal:

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

To summarize, if $f$ and $g$ are inverses and $(a, b)$ is a point on $g (x)$ (so $g (a) = b$ ), 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:

$f^{- 1} (x)$

Means inverse of $f (x)$

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

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

Examples

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 won’t explicitly say “ $g$ is the inverse of $f$ .” Instead, the clue is an equation like $f (g (x)) = x$ (or $g (f (x)) = x$ ), which tells you the functions undo each other.

1. Write the formula:

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

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

2. Find $g (2)$ :

Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (use $x$ or $b$ - either is fine):

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

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

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it follows that $g (2) = 1$ . Substitute into the derivative formula:

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

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

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

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

Evaluate at $x = 1$ :

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

$= 3$

Compute $g^{'} (a)$

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

From step 2, the point of tangency on $g$ is $(2, 1)$ . Use point-slope form:

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

To verify this in Desmos, type the equations

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

This is $f (x)$

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

This is $g (x)$ , the inverse

Equation of tangent line

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

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 $g (5)$ :

Because $f$ and $g$ are inverses and $f (2) = 5$ , you can swap the inputs/outputs:

$g (5) = 2$

Substitute into the formula:

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

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

The problem gives $f^{'} (2) = 7$ .

Compute $g^{'} (a)$

Take the reciprocal:

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

Table problem

You may also encounter problems where you extract values from a table.

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,

$g (15) = b ⟺ f (b) = 15$

Look for where $f (x) = 15$ in the table. That happens 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), $g (15) = 3$ . Substitute:

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

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

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

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

Take 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)

Use the formula:

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

Because $g$ and $f$ are inverses,

$g (6) = b ⟺ f (b) = 6$

From the table, $f (x) = 6$ when $x = 3$ , so $g (6) = 3$ . Substitute:

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

From the table, $f^{'} (3) = 4$ . Therefore,

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

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

(spoiler)

Here you work backward from the inverse-derivative formula:

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

If $g^{'} (2) = \frac{1}{8}$ , then the denominator must be $8$ , so

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

Next, find $g (2)$ . Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (1) = 2$ , so $g (2) = 1$ . Substitute into $f^{'} (g (2)) = 8$ :

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

Therefore,

$f^{'} (1) = 8$

Derivatives of 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, you would swap $x$ and $y$ :

$x = y^{5} + y$

Then you’d solve for $y$ , which would give $f^{- 1} (x)$ .

In many problems, explicitly solving for the inverse function is more work than you need. Instead, you can use a formula that gives the derivative of the inverse at a specific input value.

Given $f (x)$ , we start from the defining property of inverse functions:

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

Differentiate both sides with respect to $x$ :

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

Now apply the chain rule to the left side:

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

Divide both sides by $f^{'} (f^{- 1} (x))$ :

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

It can be easier to read if you rename the inverse function. Let $g (x) = f^{- 1} (x)$ . Then the same formula becomes

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

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 will define a function $f (x)$ and ask for the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , which is $g^{'} (a)$ . 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 $g^{'} (a)$ (here, $g^{'} (8)$ ):

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 $g (8)$ :

Focus on the inside of the denominator first. Since $f$ and $g$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set $f (b)$ equal to $8$ and solve for $b$ :

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

That means

$f (2) = 8 ⟺ g (8) = 2$

Substitute $g (8) = 2$ into the formula:

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

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

Differentiate $f (x) = x^{3}$ :

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

Evaluate at $x = 2$ :

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

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

Now take the reciprocal:

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

To summarize, if $f$ and $g$ are inverses and $(a, b)$ is a point on $g (x)$ (so $g (a) = b$ ), 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:

$f^{- 1} (x)$

Means inverse of $f (x)$

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

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

Examples

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)

1. Write the formula:

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

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

2. Find $g (2)$ :

Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (use $x$ or $b$ - either is fine):

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

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

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it follows that $g (2) = 1$ . Substitute into the derivative formula:

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

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

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

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

Evaluate at $x = 1$ :

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

$= 3$

Compute $g^{'} (a)$

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

From step 2, the point of tangency on $g$ is $(2, 1)$ . Use point-slope form:

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

To verify this in Desmos, type the equations

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

This is $f (x)$

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

This is $g (x)$ , the inverse

Equation of tangent line

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

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 $g (5)$ :

Because $f$ and $g$ are inverses and $f (2) = 5$ , you can swap the inputs/outputs:

$g (5) = 2$

Substitute into the formula:

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

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

The problem gives $f^{'} (2) = 7$ .

Compute $g^{'} (a)$

Take the reciprocal:

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

Table problem

You may also encounter problems where you extract values from a table.

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,

$g (15) = b ⟺ f (b) = 15$

Look for where $f (x) = 15$ in the table. That happens 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), $g (15) = 3$ . Substitute:

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

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

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

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

Take 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)

Use the formula:

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

Because $g$ and $f$ are inverses,

$g (6) = b ⟺ f (b) = 6$

From the table, $f (x) = 6$ when $x = 3$ , so $g (6) = 3$ . Substitute:

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

From the table, $f^{'} (3) = 4$ . Therefore,

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

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

(spoiler)

Here you work backward from the inverse-derivative formula:

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

If $g^{'} (2) = \frac{1}{8}$ , then the denominator must be $8$ , so

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

Next, find $g (2)$ . Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (1) = 2$ , so $g (2) = 1$ . Substitute into $f^{'} (g (2)) = 8$ :

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

Therefore,

$f^{'} (1) = 8$

Achievable AP Calculus AB

3. Advanced differentiation

Our AP Calculus AB course is currently in development and is a work-in-progress.

Derivatives of inverse functions

7 min read

Font

Discuss

Feedback

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, you would swap $x$ and $y$ :

$x = y^{5} + y$

Then you’d solve for $y$ , which would give $f^{- 1} (x)$ .

In many problems, explicitly solving for the inverse function is more work than you need. Instead, you can use a formula that gives the derivative of the inverse at a specific input value.

Given $f (x)$ , we start from the defining property of inverse functions:

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

Differentiate both sides with respect to $x$ :

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

Now apply the chain rule to the left side:

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

Divide both sides by $f^{'} (f^{- 1} (x))$ :

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

It can be easier to read if you rename the inverse function. Let $g (x) = f^{- 1} (x)$ . Then the same formula becomes

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

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 will define a function $f (x)$ and ask for the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , which is $g^{'} (a)$ . 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 $g^{'} (a)$ (here, $g^{'} (8)$ ):

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 $g (8)$ :

Focus on the inside of the denominator first. Since $f$ and $g$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set $f (b)$ equal to $8$ and solve for $b$ :

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

That means

$f (2) = 8 ⟺ g (8) = 2$

Substitute $g (8) = 2$ into the formula:

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

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

Differentiate $f (x) = x^{3}$ :

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

Evaluate at $x = 2$ :

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

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

Now take the reciprocal:

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

To summarize, if $f$ and $g$ are inverses and $(a, b)$ is a point on $g (x)$ (so $g (a) = b$ ), 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:

$f^{- 1} (x)$

Means inverse of $f (x)$

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

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

Examples

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)

1. Write the formula:

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

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

2. Find $g (2)$ :

Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (use $x$ or $b$ - either is fine):

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

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

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it follows that $g (2) = 1$ . Substitute into the derivative formula:

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

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

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

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

Evaluate at $x = 1$ :

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

$= 3$

Compute $g^{'} (a)$

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

From step 2, the point of tangency on $g$ is $(2, 1)$ . Use point-slope form:

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

To verify this in Desmos, type the equations

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

This is $f (x)$

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

This is $g (x)$ , the inverse

Equation of tangent line

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

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 $g (5)$ :

Because $f$ and $g$ are inverses and $f (2) = 5$ , you can swap the inputs/outputs:

$g (5) = 2$

Substitute into the formula:

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

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

The problem gives $f^{'} (2) = 7$ .

Compute $g^{'} (a)$

Take the reciprocal:

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

Table problem

You may also encounter problems where you extract values from a table.

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,

$g (15) = b ⟺ f (b) = 15$

Look for where $f (x) = 15$ in the table. That happens 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), $g (15) = 3$ . Substitute:

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

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

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

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

Take 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)

Use the formula:

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

Because $g$ and $f$ are inverses,

$g (6) = b ⟺ f (b) = 6$

From the table, $f (x) = 6$ when $x = 3$ , so $g (6) = 3$ . Substitute:

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

From the table, $f^{'} (3) = 4$ . Therefore,

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

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

(spoiler)

Here you work backward from the inverse-derivative formula:

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

If $g^{'} (2) = \frac{1}{8}$ , then the denominator must be $8$ , so

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

Next, find $g (2)$ . Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (1) = 2$ , so $g (2) = 1$ . Substitute into $f^{'} (g (2)) = 8$ :

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

Therefore,

$f^{'} (1) = 8$

Derivatives of 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, you would swap $x$ and $y$ :

$x = y^{5} + y$

Then you’d solve for $y$ , which would give $f^{- 1} (x)$ .

In many problems, explicitly solving for the inverse function is more work than you need. Instead, you can use a formula that gives the derivative of the inverse at a specific input value.

Given $f (x)$ , we start from the defining property of inverse functions:

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

Differentiate both sides with respect to $x$ :

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

Now apply the chain rule to the left side:

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

Divide both sides by $f^{'} (f^{- 1} (x))$ :

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

It can be easier to read if you rename the inverse function. Let $g (x) = f^{- 1} (x)$ . Then the same formula becomes

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

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 will define a function $f (x)$ and ask for the slope of the tangent line to the inverse function $g (x)$ at $x = a$ , which is $g^{'} (a)$ . 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 $g^{'} (a)$ (here, $g^{'} (8)$ ):

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 $g (8)$ :

Focus on the inside of the denominator first. Since $f$ and $g$ are inverses,

$g (8) = b ⟺ f (b) = 8$

So set $f (b)$ equal to $8$ and solve for $b$ :

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

That means

$f (2) = 8 ⟺ g (8) = 2$

Substitute $g (8) = 2$ into the formula:

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

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

Differentiate $f (x) = x^{3}$ :

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

Evaluate at $x = 2$ :

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

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

Now take the reciprocal:

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

To summarize, if $f$ and $g$ are inverses and $(a, b)$ is a point on $g (x)$ (so $g (a) = b$ ), 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:

$f^{- 1} (x)$

Means inverse of $f (x)$

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

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

Examples

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)

1. Write the formula:

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

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

2. Find $g (2)$ :

Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

So set $f (x) = x^{3} - x^{2} + 2 x$ equal to $2$ (use $x$ or $b$ - either is fine):

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

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

Factor by grouping:

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

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

$x = 1$

Since $f (1) = 2$ , it follows that $g (2) = 1$ . Substitute into the derivative formula:

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

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

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

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

Evaluate at $x = 1$ :

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

$= 3$

Compute $g^{'} (a)$

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

From step 2, the point of tangency on $g$ is $(2, 1)$ . Use point-slope form:

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

To verify this in Desmos, type the equations

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

This is $f (x)$

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

This is $g (x)$ , the inverse

Equation of tangent line

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

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 $g (5)$ :

Because $f$ and $g$ are inverses and $f (2) = 5$ , you can swap the inputs/outputs:

$g (5) = 2$

Substitute into the formula:

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

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

The problem gives $f^{'} (2) = 7$ .

Compute $g^{'} (a)$

Take the reciprocal:

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

Table problem

You may also encounter problems where you extract values from a table.

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,

$g (15) = b ⟺ f (b) = 15$

Look for where $f (x) = 15$ in the table. That happens 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), $g (15) = 3$ . Substitute:

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

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

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

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

Take 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)

Use the formula:

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

Because $g$ and $f$ are inverses,

$g (6) = b ⟺ f (b) = 6$

From the table, $f (x) = 6$ when $x = 3$ , so $g (6) = 3$ . Substitute:

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

From the table, $f^{'} (3) = 4$ . Therefore,

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

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

(spoiler)

Here you work backward from the inverse-derivative formula:

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

If $g^{'} (2) = \frac{1}{8}$ , then the denominator must be $8$ , so

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

Next, find $g (2)$ . Since $g$ and $f$ are inverses,

$g (2) = b ⟺ f (b) = 2$

From the table, $f (1) = 2$ , so $g (2) = 1$ . Substitute into $f^{'} (g (2)) = 8$ :

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

Therefore,

$f^{'} (1) = 8$