Derivatives of Inverse Functions

If $f$ is differentiable and strictly monotone, its inverse function $g = f^{- 1}$ is also differentiable. The derivative of the inverse is the reciprocal of the derivative of the original function, evaluated at the corresponding point.

Notation	Formula
Function form	(f^{-1})'(y)=\dfrac{1}{f'(f^{-1}(y))}
Leibniz form	$\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}}$
At a point	${(\frac{d x}{d y})}_{y_{0}} = \frac{1}{{(\frac{d y}{d x})}_{x_{0}}}$

When Does the Inverse Have a Derivative?

If f'(x)>0 (or f'(x)<0) everywhere on $(a, b)$ , then:

The graph of $f$ always rises (or always falls), so $f$ is one-to-one and has an inverse $g$ .
The graph of $f$ has no corners or cusps (since f' exists), so neither does the graph of $g$ .
The tangent to $f$ is never horizontal (since f'\neq0), so the tangent to $g$ is never vertical.

Therefore $g = f^{- 1}$ is differentiable.

Intuitive Derivation

Since $y = f (x) \Leftrightarrow x = f^{- 1} (y) = g (y)$ :

\frac{Δ x}{Δ y} = \frac{1}{\frac{Δ y}{Δ x}}

As $Δ x \to 0$ , $Δ y \to 0$ as well (by continuity), so

g'(y)=\lim_{\Delta y\to0}\frac{\Delta x}{\Delta y}=\frac{1}{\displaystyle\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}}=\frac{1}{f'(x)}

**Graphic approach.** Consider a point

(a, f (a))

on the graph of

f

. The tangent line

L

has equation y-f(a)=f'(a)(x-a).

Graph of f and its tangent line L at point (a, f(a))

Reflecting through

y = x

maps

f

f^{- 1}

and maps

L

to the tangent L' through

(f (a), a)

. Interchanging

x

and

y

in the equation of

L

: L':\quad y-a=\frac{1}{f'(a)}(x-f(a)) The slope of L' is 1/f'(a). Since L' is tangent to

y = f^{- 1} (x)

(f (a), a)

: \boxed{(f^{-1})'(f(a))=\frac{1}{f'(a)}} Setting

b = f (a)

(so

a = f^{- 1} (b)

): (f^{-1})'(b)=\dfrac{1}{f'(f^{-1}(b))}.

Graph of f^{-1} and reflected tangent line L'

The Theorem

The Derivative Rule for Inverses. Assume $f$ is differentiable on $(a, b)$ with f'(x)>0 (or f'(x)<0) for all $x \in (a, b)$ . Then the inverse $g = f^{- 1}$ exists and is differentiable, with

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

or equivalently, g'=\dfrac{1}{f'\circ g}.

In Leibniz notation:

\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}}

Alternative derivation using implicit differentiation. Start from $g (y) = x$ and differentiate with respect to $x$ :

\frac{d}{dx}(g(y))=1\quad\Rightarrow\quad g'(y)\frac{dy}{dx}=1\quad\Rightarrow\quad g'(y)=\frac{1}{f'(x)}

Example check. For $f (x) = x^{2}$ ( $x > 0$ ), the inverse is $g (y) = \sqrt{y}$ . Since f'(x)=2x:

g'(y)=\frac{1}{2x}=\frac{1}{2\sqrt{y}}\qquad(x=\sqrt{y})

Replacing $y$ by $x$ : (f^{-1})'(x)=\dfrac{1}{2\sqrt{x}}.

Worked Examples

Let $f (x) = 2 x^{3} + 3 x^{2} + 6 x + 1$ . Find (f^{-1})'(1).

Solution

f'(x)=6x^2+6x+6=6[(x+1/2)^2+3/4]>0, so

f^{- 1}

exists and is differentiable everywhere. By the Derivative Rule for Inverses: (f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))} To find

f^{- 1} (1)

: solve

f (x) = 1

, i.e.,

2 x^{3} + 3 x^{2} + 6 x + 1 = 1

, i.e.,

x (2 x^{2} + 3 x + 6) = 0

. Only

x = 0

is real. So

f^{- 1} (1) = 0

. f'(0)=6, therefore (f^{-1})'(1)=\dfrac{1}{6}.

Let $y = f (x) = x^{2} + 2 x$ for $x > - 1$ . Find $\frac{d}{d x} f^{- 1} (x)$ .

Solution

**Method (a):** Find

f^{- 1}

explicitly. From

y = x^{2} + 2 x

x = - 1 + \sqrt{1 + y} (x > - 1)

f^{- 1} (x) = - 1 + \sqrt{1 + x}

and

\frac{d}{d x} f^{- 1} (x) = \frac{1}{2 \sqrt{1 + x}}

. **Method (b):** Use the Derivative Rule. f'(x)=2x+2, so

\frac{d x}{d y} = \frac{1}{2 x + 2} = \frac{1}{2 (- 1 + \sqrt{1 + y}) + 2} = \frac{1}{2 \sqrt{1 + y}}

Replacing

y

x

: (f^{-1})'(x)=\dfrac{1}{2\sqrt{1+x}}.

Suppose $h (x) = \tan (f^{- 1} (x))$ , and we know $f (π / 4) = 1$ and f'(\pi/4)=5. Find h'(1).

Solution

Let

u = f^{- 1} (x)

. By the Chain Rule: h'(x)=(f^{-1})'(x)\cdot\sec^2(f^{-1}(x))=\frac{1+\tan^2(f^{-1}(x))}{f'(f^{-1}(x))} Since

f (π / 4) = 1

means

f^{- 1} (1) = π / 4

: h'(1)=\frac{1+\tan^2(\pi/4)}{f'(\pi/4)}=\frac{1+1}{5}=\frac{2}{5}

Let $f (x) = 4 x^{3} - 3 x^{2} + x + 1$ and $g = f^{- 1}$ . Find g'(1).

Solution

We use g'(1)=\dfrac{1}{f'(g(1))}. **Step 1:** Find

g (1)

. Solve

f (x) = 1

4 x^{3} - 3 x^{2} + x + 1 = 1 \Rightarrow x (4 x^{2} - 3 x + 1) = 0

The discriminant of

4 x^{2} - 3 x + 1

9 - 16 = - 7 < 0

, so the quadratic has no real roots. The only solution is

x = 0

, so

g (1) = 0

. **Step 2:** Compute f'(0). f'(x)=12x^2-6x+2, so f'(0)=2. **Step 3:** g'(1)=\dfrac{1}{f'(0)}=\dfrac{1}{2}.

Frequently Asked Questions

Why does the derivative of

f^{- 1}

involve f' evaluated at

f^{- 1} (y)

The derivative rule connects the slope of

f^{- 1}

y

to the slope of

f

at the corresponding

x

-value, which is

f^{- 1} (y)

. Geometrically, reflecting a curve through

y = x

turns a slope of

m

into a slope of

1 / m

What if f'(x)=0 somewhere?

If f'(x_0)=0, then the tangent to

f

x_{0}

is horizontal. Reflecting that through

y = x

gives a vertical tangent, meaning

f^{- 1}

is not differentiable at the corresponding point.

Do I always need to find an explicit formula for

f^{- 1}

to compute its derivative?

No. The Derivative Rule for Inverses lets you compute (f^{-1})'(y)=1/f'(f^{-1}(y)) without an explicit formula, as long as you can find the

x

where

f (x) = y

(i.e.,

f^{- 1} (y)

) and compute f' there.

When Does the Inverse Have a Derivative?

Intuitive Derivation

The Theorem

Worked Examples

Frequently Asked Questions

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