Lecture #17: Inverse Functions & More Logarithmic Derivatives

MATH 145 – Calculus I

Torin Quinlivan
tquinlivan@knox.edu

Knox College

October 27, 2025

Derivatives of the form \(f(x)^{g(x)}\)

We know how to find \(\dfrac{d}{dx} x^a\) when \(a\) is a constant: \[ \dfrac{d}{dx} x^a = a x^{a-1} \]

We also know now how to find \(\dfrac{d}{dx} a^x\) when \(a\) is a constant: \[ \dfrac{d}{dx} a^x = a^x \ln(a) \]

Derivatives of the form \(f(x)^{g(x)}\)

What is the derivative of \(\dfrac{d}{dx} x^x\)?

\[ \begin{aligned} \dfrac{d}{dx} x^x &= \dfrac{d}{dx} e^{\ln(x^x)} \\ &= e^{x \ln(x)} \\ &= e^{x \ln(x)} \left( \frac{d}{dx} \left( x \ln(x) \right) \right) \\ &= e^{\ln(x^x)} \left( x \left( \frac{d}{dx} \left( \ln(x) \right) \right) + \ln(x) \left( \frac{d}{dx} x \right) \right) \\ &= x^x \left( x \left( \frac{1}{x} \right) + \left( \ln(x) \right) (1) \right) \\ &= x^x ( \ln(x) + 1) \end{aligned} \]

Example: Derivatives of the form \(f(x)^{g(x)}\)

Examples:

  1. Find \(\dfrac{d}{dx} (x^2 + 1)^{(x^2 + 1)}\).
  2. Find \(\dfrac{d}{dx} \sin(x)^{\sin(x)}\).

Derivatives of the form \(f(x)^{g(x)}\)

What is the derivative of \(\dfrac{d}{dx} f(x)^{g(x)}\)?

\[ \begin{aligned} \dfrac{d}{dx} f(x)^{g(x)} &= \dfrac{d}{dx} e^{\ln(f(x)^{g(x))}} \\ &= e^{g(x) \ln(f(x))} \\ &= e^{g(x) \ln(f(x))} \left( \frac{d}{dx} \left( g(x) \ln(f(x)) \right) \right) \\ &= e^{\ln(f(x)^{g(x))}} \left( g(x) \left( \frac{d}{dx} \left( \ln(f(x)) \right) \right) + \ln(f(x)) \left( \frac{d}{dx} g(x) \right) \right) \\ &= f(x)^{g(x)} \left( g(x) \left( \frac{f'(x)}{f(x)} \right) + \left( \ln(f(x)) \right) g'(x) \right) \\ &= f(x)^{g(x)} \left( g(x) \left( \frac{f'(x)}{f(x)} \right) + g'(x) \ln(f(x)) \right) \end{aligned} \]

Example: Derivatives of the form \(f(x)^{g(x)}\)

Examples:

  1. Find \(\dfrac{d}{dx} (x^2 + 1)^{x^3}\).
  2. Find \(\dfrac{d}{dx} \sin(x)^{\cos(x)}\).

Inverse Functions

Given a function \(f: A \to B\), the inverse function of \(f(x)\) is the function \(f^{-1} : B \to A\) such that \[ \begin{aligned} f(f^{-1}(x)) &= x & &\textrm{and} & f^{-1}(f(x)) = x \end{aligned} \]

Inverse Functions

Only one-to-one functions, functions for which if \(f(x) = f(y)\), then \(x = y\), for all \(x\) and \(y\) in the domain.

One-to-one functions must pass the horizontal line test.

If a function is not one-to-one, you can sometimes restrict the domain to get a one-to-one function.

The graph of the inverse function \(f^{-1}\) is the graph of \(f\) reflected over the line \(y = x\).

Inverse Trigonometric Functions

By restricting the domain, the trigonometric functions can be made to have inverse functions.

For instance, if we restrict the domain of the sine function to \([-\frac{\pi}{2}, \frac{\pi}{2}]\), then we can define the inverse function \(\sin^{-1} : [-1, 1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]\), also known as \(\arcsin(x)\).

Similarly, if we restrict the domain of the tangent function to \([-\frac{\pi}{2}, \frac{\pi}{2}]\), the we can define the inverse function \(\tan^{-1} : [-\infty, \infty] \to [-\frac{\pi}{2}, \frac{\pi}{2}]\), also known as \(\arctan(x)\).

The derivatives of these inverse functions are very important in the study of antiderivatives because \(\sin^2(x) + \cos^2(x) = 1\).

Inverse Trigonometric Functions

To find the derivative of \(\sin^{-1}\), we will use a trick built around the fact that \(\sin(\sin^{-1}(x)) = x\), and the fact that \(\sin^2(x) + \cos^2(x) = 1\) implies \(\cos(x) = \sqrt{1 - \sin^2(x)}\).

\[ \begin{aligned} \frac{d}{dx} x &= 1 \\ \frac{d}{dx} \sin(\sin^{-1}(x)) &= 1 \\ \cos(\sin^{-1}(x)) \left( \frac{d}{dx} \sin^{-1}(x) \right) &= 1 \\ \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\cos(\sin^{-1}(x))} \\ \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\sqrt{1 - (\sin(\sin^{-1}(x)))^2}} \\ \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\sqrt{1 - x^2}} \end{aligned} \]

Example: Inverse Trigonometric Functions

Examples:

  1. Find \(\dfrac{d}{dx} \cos^{-1}(x)\).
  2. Find \(\dfrac{d}{dx} \tan^{-1}(x)\).
  3. Find \(\dfrac{d}{dx} \sqrt{x} \tan^{-1}(x^2)\).
  4. Find \(\dfrac{d}{dx} e^{\sin^{-1}(\ln(x))}\).

Derivatives of Inverse Trigonometric Functions

  • \(\dfrac{d}{dx} \sin^{-1}(x) = \dfrac{1}{\sqrt{1 - x^2}}\).
  • \(\dfrac{d}{dx} \cos^{-1}(x) = -\dfrac{1}{\sqrt{1 - x^2}}\).
  • \(\dfrac{d}{dx} \tan^{-1}(x) = \dfrac{1}{1 + x^2}\).