3.5 Derivatives of inverse functions

Achievable AP Calculus AB

3. Advanced differentiation

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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 $f (x) = x^{5}$ . To find its inverse, you would swap $x$ and $y$ :

$x = y^{5}$

Then solving for $y$ gives the inverse function

$f^{- 1} (x) = x^{1/5}$

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 $f^{- 1} (5) = 3$ .

Explicitly solving for the inverse function is not always possible. Instead, a formula is used to find the derivative of the inverse function 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:

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

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

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

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 $f (x),$ or $\frac{1}{f ( x )}$

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

Derivative of an inverse function:

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

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

Examples

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 $x = 8$ .

(spoiler)

Follow these steps to find $g^{'} (8)$ , the slope of the tangent line to $g$ at $x = 8$ .

1. Write the formula:

The derivative of the inverse function at $x = 8$ is

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

2. Find $g (8)$ :

Since $f$ and $g$ are inverses, inputs and outputs are swapped.

$g (8) = b ⟺ f (b) = 8$

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

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

This 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 $g^{'} (8)$ :

Now take the reciprocal:

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

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

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

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

Find the equation of the tangent line to $g (x)$ at $x = 2$ if $f (x) = x^{3} - x^{2} + 2 x$ .

(spoiler)

Sometimes a problem won’t explicitly state that $g$ 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 two functions are inverses.

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 $x = 1$ :

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

4. Compute $g^{'} (2)$ :

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

From step 2, the point $(2, 1)$ is on $g$ . So the equation of the tangent line to $g$ is

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

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
The tangent line equation - verify it is tangent to $g (x)$ at $(2, 1)$

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, 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)

Working backwards fom 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 $f (x) = x^{5}$ . To find its inverse, you would swap $x$ and $y$ :

$x = y^{5}$

Then solving for $y$ gives the inverse function

$f^{- 1} (x) = x^{1/5}$

Explicitly solving for the inverse function is not always possible. Instead, a formula is used to find the derivative of the inverse function 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:

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

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

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

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 $f (x),$ or $\frac{1}{f ( x )}$

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

Derivative of an inverse function:

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

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

Examples

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 $x = 8$ .

(spoiler)

Follow these steps to find $g^{'} (8)$ , the slope of the tangent line to $g$ at $x = 8$ .

1. Write the formula:

The derivative of the inverse function at $x = 8$ is

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

2. Find $g (8)$ :

Since $f$ and $g$ are inverses, inputs and outputs are swapped.

$g (8) = b ⟺ f (b) = 8$

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

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

This 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 $g^{'} (8)$ :

Now take the reciprocal:

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

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

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

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

Find the equation of the tangent line to $g (x)$ at $x = 2$ if $f (x) = x^{3} - x^{2} + 2 x$ .

(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 $x = 1$ :

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

4. Compute $g^{'} (2)$ :

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

From step 2, the point $(2, 1)$ is on $g$ . So the equation of the tangent line to $g$ is

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

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
The tangent line equation - verify it is tangent to $g (x)$ at $(2, 1)$

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, 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)

Working backwards fom 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 now in "early access" - get 50% off for a limited time.