Derivatives of Inverse Functions

107. Derivatives of Inverse Functions

Finding the derivative of the inverse function $𝑓^{- 1} (𝑥)$ at a given point directly from the function $𝑓 (𝑥)$ . Instead of explicitly computing the inverse function $𝑓^{- 1} (𝑥)$ , we use the inverse function derivative formula:

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

This approach allows us to determine the derivative of the inverse function without needing to express $𝑓^{- 1} (𝑥)$ explicitly. Instead, we find the value of $𝑥$ that satisfies $𝑓 (𝑥) = 𝑎$ (where a a is the given point), evaluate $𝑓^{'} (𝑥)$ , and apply the formula.

1. Definition of Inverse Function

\begin{matrix} 𝑓 (𝑓^{- 1} (𝑥)) = 𝑥 \\ 𝑓^{- 1} (𝑓 (𝑥)) = 𝑥 \end{matrix}

2. Differentiate Both Sides

Differenatiate both sides with respect to $𝑥$

\frac{𝑑}{𝑑 𝑥} [𝑓 (𝑓^{- 1} (𝑥))] = \frac{𝑑}{𝑑 𝑥} [𝑥]

The right-hand side simplifies to:

\frac{𝑑}{𝑑 𝑥} [𝑓 (𝑓^{- 1} (𝑥))] = 1

Using chain-rule:

\frac{𝑑}{𝑑 𝑥} [𝑓 (𝑔 (𝑥))] = 𝑓^{'} (𝑔 (𝑥)) \cdot 𝑔^{'} (𝑥)

The left side expands as:

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

This we get:

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

3. Solve for $\frac{𝑑}{𝑑 𝑥} [𝑓^{- 1} (𝑥)]$

Rearrange to isolate $\frac{𝑑}{𝑑 𝑥} [𝑓^{- 1} (𝑥)]$ :

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

Example

Given the function:

𝑓 (𝑥) = 𝑥^{3}

We want to find $\frac{𝑑}{𝑑 𝑥} 𝑓^{- 1} (𝑥)$ at $𝑥 = 0.5$ using the inverse function derivative formula:

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

1. Compute $𝑓^{'} (𝑥)$

Differenatiate $𝑓 (𝑥)$ :

𝑓^{'} (𝑥) = 3 𝑥^{2}

2. Solve for $𝑥$ such that $𝑓 (𝑥) = 0.5$

We need to find $𝑥$ such that:

𝑥^{3} = 0.5

Solving for $𝑥$ :

𝑥 = \sqrt[3]{0.5}

3. Compute $𝑓^{'} (\sqrt[3]{0.5})$

Evaluate the derivative at $𝑥 = \sqrt[3]{0.5}$ :

𝑓^{'} (\sqrt[3]{0.5}) = 3 {(\sqrt[3]{0.5})}^{2}

4. Use the formula

\frac{𝑑}{𝑑 𝑥} 𝑓^{- 1} (0.5) = \frac{1}{3 {(\sqrt[3]{0.5})}^{2}}

5. Interpretation

The expression:

\frac{𝑑}{𝑑 𝑥} 𝑓^{- 1} (0.5) = \frac{1}{3 {(\sqrt[3]{0.5})}^{2}}

represents the derivative of the inverse function $𝑓^{- 1} (𝑥)$ evaluated at $𝑥 = 0.5$ . This means it gives the slope of the tangent line to the inverse function at $𝑥 = 0.5$ .

inverse.py

from sympy import symbols, solve

# y = x**3 + x

x, y = symbols('x y')

f = x**3 + x - y

inverse = solve(f, x)

print(inverse)

107.0.1. Derivative Inverse Sin

\frac{𝑑}{𝑑 𝑥} [{sin}^{- 1} (𝑥)] = \frac{1}{\sqrt{1 - 𝑥^{2}}}

\frac{𝑑}{𝑑 𝑥} [arcsin (𝑥)] = \frac{1}{\sqrt{1 - 𝑥^{2}}}

107.0.2. Derivative Inverse Cos

\frac{𝑑}{𝑑 𝑥} [{cos}^{- 1} (𝑥)] = - \frac{1}{\sqrt{1 - 𝑥^{2}}}

\frac{𝑑}{𝑑 𝑥} [arccos (𝑥)] = - \frac{1}{\sqrt{1 - 𝑥^{2}}}

107.0.3. Derivative Inverse Tan

\frac{𝑑}{𝑑 𝑥} [{tan}^{- 1} (𝑥)] = - \frac{1}{1 + 𝑥^{2}}

\frac{𝑑}{𝑑 𝑥} [arctan (𝑥)] = - \frac{1}{1 + 𝑥^{2}}