Differentiating Inverse Functions

Differentiating inverse functions allows us to compute the derivative of an inverse function \( g(x) \), even when we cannot explicitly find its expression.

Let \( f \) be an invertible function, meaning \( f: A \to B \) is both one-to-one and onto. If \( f \) is invertible, then there exists a function \( g \) such that

\[ g(f(x)) = x \quad \text{and} \quad f(g(x)) = x. \]

We call \( g \) the inverse of \( f \) and write it as \( g = f^{-1} \). While some functions have explicit inverses, others do not. For example,

\[ y = x + \sin x \]

is invertible, but we cannot write its inverse in a closed form. However, we can find the derivative of the inverse function directly.

Since \( f(g(x)) = x \), differentiating both sides,

\[ f'(g(x)) \cdot g'(x) = 1. \]

Solving for \( g'(x) \),

\[ g'(x) = \frac{1}{f'(g(x))}. \]

Thus, the derivative of the inverse function is computed using the formula

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

Example

Let \( f(x) = x^3 + x - 1 \), and suppose \( g \) is its inverse. Find \( g'(-1) \).

By the inverse derivative formula,

\[ g'(-1) = \frac{1}{f'(g(-1))}. \]

At this point, we need to determine \( g(-1) \), meaning we need to find \( x \) such that

\[ f(x) = -1. \]

Trying \( x = 0 \),

\[ f(0) = 0^3 + 0 - 1 = -1. \]

Since \( f(0) = -1 \), and \( g \) is the inverse function, this implies

\[ g(-1) = 0. \]

Thus,

\[ g'(-1) = \frac{1}{f'(0)}. \]

Now, computing \( f'(x) \),

\[ f'(x) = 3x^2 + 1. \]

Substituting \( x = 0 \),

\[ f'(0) = 3(0)^2 + 1 = 1. \]

\[ g'(-1) = \frac{1}{1} = 1. \]

Thus,

\[ g'(-1) = 1. \]

Let \( f \) be an invertible function, and let \( g \) be its inverse. This means that if \( y = f(x) \), then applying the inverse function gives

\[ g(y) = x. \]

Since \( g \) and \( f \) are inverses, we have the identity

\[ g(f(x)) = x. \]

Differentiating both sides with respect to \( x \),

\[ g'(f(x)) f'(x) = 1. \]

Solving for \( g'(f(x)) \), we obtain

\[ g'(f(x)) = \frac{1}{f'(x)}. \]

This provides another way to express the derivative of an inverse function. Since we assumed \( y = f(x) \), we substitute \( y \) in place of \( f(x) \),

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

Now, expressing this in Leibniz Notation,

\[ \frac{dg(y)}{dy} = \frac{1}{\frac{d f(x)}{dx}}. \]

Rewriting,

\[ \boxed{\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}}. \]