How do you find the derivative of the inverse at a?

Find b with f of b equals a, compute f prime of b, then the answer is one over f prime of b.

What if f prime of b is zero?

Then the inverse has a vertical tangent there and its derivative does not exist.

How do you find the derivative of the inverse at a?

Find b with f of b equals a, compute f prime of b, then the answer is one over f prime of b.

What if f prime of b is zero?

Then the inverse has a vertical tangent there and its derivative does not exist.

Year 12 Maths Extension 1 (2027) Calculus

Derivatives of inverse functions

Q: What is the derivative of an inverse function?

It is one over f prime evaluated at the inverse of x; that is, one over f prime at the pre-image.

Q: Why are the gradients reciprocals?

Reflecting in y equals x swaps rise and run, so a gradient m becomes one over m at the reflected point.

Q: Do you need a formula for the inverse?

No. You only need the pre-image b and the value f prime of b.

20 practice questions 2 video lessons Theory + worked examples

Understand derivatives of inverse functions in NSW Year 12 Mathematics Extension 1. The gradient of an inverse function at a point is the reciprocal of the original function's gradient at the matching point.

You will learn why this reciprocal relationship holds, apply it to find gradients and tangents without first finding the inverse, and connect it to the chain rule — a neat calculus result used throughout Extension 1.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Derivative of Inverse Functions Examples & Practice Problems - Calculus Watch
Finding the Derivative of an Inverse Function - Calculus I Watch

Theory

The inverse function $f^{- 1}$ reflects $f$ in $y = x$ , so their tangent gradients are reciprocals. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) uses $(f^{- 1})^{'} (x) = \frac{1}{f^{'} (f^{- 1} (x))}$ to differentiate an inverse without a formula for it.

The inverse function $f^{- 1}$ undoes $f$ : if $(a, b)$ lies on $y = f (x)$ , then $(b, a)$ lies on $y = f^{- 1} (x)$ — a reflection in the line $y = x$ .

Reflecting swaps run and rise, so tangents at reflected points have reciprocal gradients: a gradient $m$ at $(a, b)$ becomes $\frac{1}{m}$ at $(b, a)$ . This gives the derivative of the inverse without ever finding a formula for $f^{- 1}$ .

The one trap: evaluate $f^{'}$ at the pre-image $b = f^{- 1} (a)$ , not at $a$ .

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01. This is part of the Further calculus skills focus area.

f^{- 1}

reflects

f

y = x

; gradients at reflected points are reciprocals.

To find the inverse derivative at

a

: get the pre-image

b

, then take the reciprocal of the slope at

b

If $f$ is one-to-one and differentiable with $f^{'} \neq 0$ :

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

(f inverse)'(x) = 1 / f'(f inverse of x)

Equivalently, writing $y = f^{- 1} (x)$ so $x = f (y)$ :

\frac{d y}{d x} = \frac{1}{f^{'} (y)} .

dy/dx = 1 / f'(y) where x = f(y)

Reciprocal, not negative reciprocal. The reflected gradient is $\frac{1}{m}$ . If $f^{'} (b) = 0$ , the inverse has a vertical tangent there and $(f^{- 1})^{'}$ does not exist.

How to find $(f^{- 1})^{'} (a)$

Find the pre-image. Solve $f (b) = a$ for $b$ (often by inspection).
Differentiate $f$ . Compute $f^{'} (x)$ , then evaluate $f^{'} (b)$ .
Take the reciprocal: $(f^{- 1})^{'} (a) = \frac{1}{f^{'} (b)}$ .
For a tangent to $y = f^{- 1} (x)$ at $x = a$ , use the point $(a, b)$ and gradient $\frac{1}{f^{'} (b)}$ .

Example 1 — From given values

A function

f

has

f (2) = 5

and

f^{'} (2) = 8

. Find

(f^{- 1})^{'} (5)

Solution

The pre-image of $5$ is $2$ .

(f^{- 1})^{'} (5)

=

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

(f inverse)'(5) = 1/8

So $(f^{- 1})^{'} (5) = \frac{1}{8}$ .

Example 2 — Find the pre-image first

Let

f (x) = x^{3} + 4 x

. Find

(f^{- 1})^{'} (5)

Solution

$b^{3} + 4 b = 5$	$\Rightarrow$	$b = 1$
$f^{'} (x) = 3 x^{2} + 4$	$\Rightarrow$	$f^{'} (1) = 7$
$(f^{- 1})^{'} (5)$	$=$	$\frac{1}{7}$

(f inverse)'(5) = 1/7

So $(f^{- 1})^{'} (5) = \frac{1}{7}$ .

Example 3 — Tangent to the inverse

Let

f (x) = x^{3} + 5 x

, so

f (1) = 6

. Find the tangent to

y = f^{- 1} (x)

x = 6

Solution

The point is $(6, 1)$ , gradient $\frac{1}{f^{'} (1)}$ .

$f^{'} (1) = 3 + 5$	$=$	$8, m = \frac{1}{8}$
$y - 1$	$=$	$\frac{1}{8} (x - 6)$

tangent: x - 8y + 2 = 0

Tangent: $x - 8 y + 2 = 0$ .

Example 4 — Pre-image by inspection

Let

f (x) = x + e^{x}

. Find

(f^{- 1})^{'} (1)

Solution

By inspection $f (0) = 0 + 1 = 1$ , so $b = 0$ .

$f^{'} (x) = 1 + e^{x}$	$\Rightarrow$	$f^{'} (0) = 2$
$(f^{- 1})^{'} (1)$	$=$	$\frac{1}{2}$

(f inverse)'(1) = 1/2

So $(f^{- 1})^{'} (1) = \frac{1}{2}$ .

Common pitfalls

Evaluating at the wrong place.

(f^{- 1})^{'} (a) = \frac{1}{f^{'} (b)}

where

b = f^{- 1} (a)

— use the pre-image

b

, not

a

Negative reciprocal. The gradients are reciprocals

(\frac{1}{m})

, not negative reciprocals — that is for perpendicular lines.

Zero derivative. If

f^{'} (b) = 0

the inverse has a vertical tangent there, so

(f^{- 1})^{'}

does not exist.

Solving for $f^{- 1}$ . You do not need a formula for

f^{- 1}

— just the pre-image and

f^{'}

Frequently asked questions

What is the derivative of an inverse function?

$(f^{- 1})^{'} (x) = \frac{1}{f^{'} (f^{- 1} (x))}$ : the reciprocal of $f^{'}$ evaluated at the pre-image.

How do you find (f inverse)'(a)?

Find $b$ with $f (b) = a$ , compute $f^{'} (b)$ , then $(f^{- 1})^{'} (a) = \frac{1}{f^{'} (b)}$ .

Why are the gradients reciprocals?

Reflecting in $y = x$ swaps rise and run, so a gradient $m$ becomes $\frac{1}{m}$ at the reflected point.

Do you need a formula for the inverse?

No. You only need the pre-image $b$ and the value $f^{'} (b)$ .

What if f'(b) = 0?

Then the inverse has a vertical tangent at that point and its derivative does not exist there.

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Derivatives of inverse functions

📖 Theory

How to find (f−1)′(a)

Common pitfalls

Frequently asked questions

Theory

How to find $(f^{- 1})^{'} (a)$