# Derivative of Inverse Function

## Contents

## Theorem

Let $I = \closedint a b$ and $J = \closedint c d$ be closed real intervals.

Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding open real intervals.

Let $f: I \to J$ be a real function which is continuous on $I$ and differentiable on $I^o$ such that $J = f \sqbrk I$.

Let either:

- $\forall x \in I^o: D \map f x > 0$

or:

- $\forall x \in I^o: D \map f x < 0$

Then:

- $f^{-1}: J \to I$ exists and is continuous on $J$
- $f^{-1}$ is differentiable on $J^o$
- $\forall y \in J^o: D \map {f^{-1} } y = \dfrac 1 {D \map f x}$

## Proof

From Derivative of Monotone Function, it follows that $f$ is either:

- strictly increasing on $I$ (if $\forall x \in I^o: D \, \map f x > 0$)

or:

- strictly decreasing on $I$ (if $\forall x \in I^o: D \, \map f x < 0$).

Therefore from Inverse of Strictly Monotone Function it follows that $f^{-1}: J \to I$ exists.

As $f$ is continuous, from Image of Interval by Continuous Function is Interval it follows that $J$ is an interval.

By the Corollary to Limit of Increasing Function and the Corollary to Limit of Decreasing Function, $f^{-1}: J \to I$ is continuous.

Next its derivative is to be considered.

Suppose $f$ is strictly increasing.

Let $y \in J^o$.

Then $\map {f^{-1} } y \in I^o$.

Let $k = \map {f^{-1} } {y + h} - \map {f^{-1} } y$.

Thus:

- $\map {f^{-1} } {y + h} = \map {f^{-1} } y + k = x + k$

Thus:

- $y + h = \map f {x + k}$

and hence:

- $h = \map f {x + k} - y = \map f {x + k} - \map f x$

Since $f^{-1}$ is continuous on $J$, it follows that $k \to 0$ as $h \to 0$.

Also, $f^{-1}$ is strictly increasing from Inverse of Strictly Monotone Function and so $k \ne 0$ unless $h = 0$.

So by Limit of Composite Function we get:

- $\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h = \dfrac k {\map f {x + k} - \map f x}$

Thus:

- $\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h \to \dfrac 1 {D \map f x}$

as $h \to 0$.

Suppose $f$ is strictly decreasing.

Exactly the same argument applies.

$\blacksquare$

## Style Note

Leibniz's notation for derivatives $\paren {\dfrac {\d y} {\d x} }$ allows for a particularly elegant statement of this rule:

- $\dfrac {\d x} {\d y} = \dfrac 1 {\frac {\d y} {\d x} }$

where:

- $\dfrac {\d x} {\d y}$ is the derivative of $x$ with respect to $y$
- $\dfrac {\d y} {\d x}$ is the derivative of $y$ with respect to $x$.

However, do not interpret this to mean that derivatives can be treated as fractions, it is simply a convenient notation.

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 13$: General Rules of Differentiation: $13.12$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 10.15 \ (5)$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 12.10$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**inverse function**