# Derivative of an Inverse Function: Formula and Examples

> The inverse function rule: the derivative of f⁻¹ at x is 1/f'(f⁻¹(x)). The formula, why slopes are reciprocals, worked examples and arcsin and ln.

Source: https://practicecalculus.com/rules/inverse-function-rule
Updated: 2026-11-03

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

## At a glance

- **Applies to:** A one-to-one differentiable function $f$, at points where $f'(f^{-1}(x)) \ne 0$
- **Point form:** If $f(a) = b$, then $(f^{-1})'(b) = \frac{1}{f'(a)}$
- **Leibniz form:** $\frac{dx}{dy} = \frac{1}{dy/dx}$
- **Graph meaning:** Reflecting across $y = x$ swaps rise and run, so slopes become reciprocals
- **Gives:** Derivatives of $\ln x$, $\sqrt{x}$, $\arcsin x$ and $\arctan x$

## Key takeaways

- The slope of $f^{-1}$ at $(b, a)$ is the reciprocal of the slope of $f$ at $(a, b)$.
- You can find $(f^{-1})'(b)$ without a formula for $f^{-1}$: solve $f(a) = b$, then compute $\frac{1}{f'(a)}$.
- The rule fails where $f'(a) = 0$, because the inverse has a vertical tangent there.

The **inverse function rule** gives the derivative of an inverse function from the derivative of the original. If $f$ has slope $m$ at the point $(a, b)$, then $f^{-1}$ has slope $\frac{1}{m}$ at the mirror point $(b, a)$. You do not need a formula for the inverse to use it.

## The inverse function rule formula

Let $f$ be one-to-one and differentiable, with inverse $f^{-1}$. If $f'(f^{-1}(x)) \ne 0$, then $f^{-1}$ is differentiable at $x$ and

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

The **point form** is usually easier to use. If $f(a) = b$ and $f'(a) \ne 0$:

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

In Leibniz notation, with $y = f(x)$ and $x = f^{-1}(y)$, this reads $\frac{dx}{dy} = \frac{1}{dy/dx}$.

```graph
title: 'Graphs of y = x² and y = √x reflected across y = x'
caption: 'At $(2, 4)$ the parabola has slope $4$. At the mirror point $(4, 2)$ the square root has slope $\frac{1}{4}$.'
x: [0, 6]
y: [0, 6]
functions:
  - expr: x^2
    label: y = x²
    domain: [0, 2.45]
  - expr: sqrt(x)
    label: y = √x
  - expr: x
    label: y = x
    dashed: true
points:
  - [2, 4, 'slope 4']
  - [4, 2, 'slope 1/4']
```

## Why the rule works

Start from the identity $f(f^{-1}(x)) = x$ and differentiate both sides using the [chain rule](/rules/chain-rule):

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

Divide by $f'(f^{-1}(x))$ and you have the rule. Geometrically, reflecting a graph across $y = x$ swaps $x$ and $y$, which swaps rise and run, so every slope turns into its reciprocal.

## Worked examples

**Example 1: no formula for the inverse.** Let $f(x) = x^3 + x + 1$, which is increasing and therefore one-to-one. Find $(f^{-1})'(3)$.

**Step 1.** Solve $f(a) = 3$: $a^3 + a + 1 = 3$ gives $a = 1$.

**Step 2.** Differentiate: $f'(x) = 3x^2 + 1$, so $f'(1) = 4$.

**Step 3.** Take the reciprocal: $(f^{-1})'(3) = \frac{1}{4}$.

**Example 2: the derivative of $\ln x$.** Here $f(x) = e^x$, $f'(x) = e^x$ and $f^{-1}(x) = \ln x$:

$$\frac{d}{dx}\ln x = \frac{1}{e^{\ln x}} = \frac{1}{x}$$

**Example 3: the derivative of $\arcsin x$.** Here $f(x) = \sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $f'(x) = \cos x$:

$$\frac{d}{dx}\arcsin x = \frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

The last step uses $\cos\theta = \sqrt{1 - \sin^2\theta}$, which is positive on that interval. See [arcsine](/functions/arcsine).

**Example 4: the derivative of $\arctan x$.** With $f(x) = \tan x$ and $f'(x) = \sec^2 x = 1 + \tan^2 x$:

$$\frac{d}{dx}\arctan x = \frac{1}{1 + \tan^2(\arctan x)} = \frac{1}{1 + x^2}$$

## Derivatives the rule produces

| Function $f$ | Inverse $f^{-1}$ | Derivative of the inverse |
| --- | --- | --- |
| $x^2$ for $x \ge 0$ | $\sqrt{x}$ | $\frac{1}{2\sqrt{x}}$ |
| $e^x$ | $\ln x$ | $\frac{1}{x}$ |
| $\sin x$ | $\arcsin x$ | $\frac{1}{\sqrt{1 - x^2}}$ |
| $\cos x$ | $\arccos x$ | $-\frac{1}{\sqrt{1 - x^2}}$ |
| $\tan x$ | $\arctan x$ | $\frac{1}{1 + x^2}$ |

## Common mistakes

1. **Evaluating $f'$ at the wrong input.** $(f^{-1})'(b)$ uses $f'(a)$ where $f(a) = b$, not $f'(b)$. In Example 1, $f'(3) = 28$ would give the wrong answer $\frac{1}{28}$.
2. **Confusing $f^{-1}$ with $\frac{1}{f}$.** $f^{-1}$ is the inverse function, not the reciprocal. $\sin^{-1}x$ means $\arcsin x$, not $\csc x$.
3. **Ignoring where $f' = 0$.** If $f'(a) = 0$, the inverse has a vertical tangent at $b$ and is not differentiable there. For $f(x) = x^3$, the inverse $\sqrt[3]{x}$ is not differentiable at $0$.

The same result also comes out of [implicit differentiation](/rules/implicit-differentiation): write $x = f(y)$ and differentiate both sides. For more on $\ln x$ and $e^x$, read the guide to [derivatives of exponential and logarithmic functions](/blog/derivatives-of-exponential-and-logarithmic-functions).

## Frequently asked questions

### What is the formula for the derivative of an inverse function?

$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$. Equivalently, if $f(a) = b$, then $(f^{-1})'(b) = \frac{1}{f'(a)}$.

### How do you find the derivative of an inverse at a point?

Find the input $a$ with $f(a) = b$, compute $f'(a)$, and take its reciprocal. You do not need to find $f^{-1}$ itself.

### Why is the derivative of an inverse a reciprocal?

The graph of $f^{-1}$ is the graph of $f$ reflected across $y = x$. Reflection swaps rise and run, so a slope $m$ becomes $\frac{1}{m}$.

### Is the derivative of f inverse the same as 1 over f prime?

Not at the same input. $(f^{-1})'(x)$ equals $\frac{1}{f'}$ evaluated at $f^{-1}(x)$, not at $x$. Using the wrong input is the most common mistake.