Examples with detailed solutions on how to find the derivative (Differentiation) of an inverse function are presented. Also more exercises with answers are included.
Derivative of Inverse Function Formula (Theorem)
Let \( f \) be a function and \( f^{-1} \) its inverse. One of the fundamental properties of inverse functions is that:
\[ f(f^{-1}(x)) = x \]Let \( y = f^{-1}(x) \), so that:
\[ f(y) = x \]Differentiate both sides using the chain rule on the left side:
\[ \dfrac{df}{dy} \dfrac{dy}{dx} = 1 \]Solve for \( \dfrac{dy}{dx} \):
\[ \dfrac{dy}{dx} = \dfrac{df^{-1}}{dx} = \dfrac{1}{\dfrac{df}{dy}} \]which may also be written as:
where \( f' \) is the first derivative of \( f \).
Examples with Detailed Solutions
Example 1
Find the derivative of the inverse of function \( f \) given by:
\[ f(x) = \dfrac{x}{2} - 1 \]Solution to Example 1 (Click to view methods)
We present two methods to answer the above question. In the first method we calculate the inverse function and then its derivative. In the second method, we use the formula developed above.
Method 1: Explicit Inverse Function
The first method consists of finding the inverse of function \( f \) and differentiating it. To find the inverse of \( f \), we first write it as an equation:
\[ y = \dfrac{x}{2} - 1 \]Solve for \( x \):
\[ x = 2y + 2 \]Interchange \( x \) and \( y \) to obtain the inverse:
\[ y = f^{-1}(x) = 2x + 2 \]The derivative of the inverse function is:
\[ \dfrac{dy}{dx} = \dfrac{df^{-1}}{dx} = 2 \]Method 2: Using the Inverse Function Theorem Formula
Given \( f(x) = \dfrac{x}{2} - 1 \), hence:
\[ f'(x) = \dfrac{1}{2} \]Substitute \( f' \) by \( \dfrac{1}{2} \) in the formula \( \dfrac{df^{-1}}{dx} = \dfrac{1}{f'(f^{-1}(x))} \):
\[ \dfrac{df^{-1}}{dx} = \dfrac{1}{\frac{1}{2}} = 2 \]Note that the first method can be used only if we can find the inverse function explicitly.
Example 2
Find the derivative \( \dfrac{dy}{dx} \) where \( y = \arcsin x \).
Solution to Example 2 (Click to view solution)
\( \arcsin x \) is the inverse function of \( \sin x \) and hence:
\[ \sin(\arcsin(x)) = x \qquad (I) \]Given \( y = \arcsin x \). Take the sine of both sides:
\[ \sin y = \sin(\arcsin x) \]Simplify using (I):
\[ \sin y = x \]Differentiate both sides with respect to \( x \), using the chain rule on the left side:
\[ \dfrac{dy}{dx} \cos y = 1 \]Solve for \( \dfrac{dy}{dx} \):
\[ \dfrac{dy}{dx} = \dfrac{1}{\cos y} \]Make the substitution \( y = \arcsin x \):
\[ \dfrac{dy}{dx} = \dfrac{1}{\cos(\arcsin x)} \qquad (II) \]Simplify \( \cos(\arcsin x) \) using the trigonometric identity \( \cos x = \sqrt{1 - \sin^2 x} \):
\[ \cos(\arcsin x) = \sqrt{1 - \sin^2(\arcsin x)} \]Simplify using the property of inverse functions \( \sin(\arcsin x) = x \):
\[ \cos(\arcsin x) = \sqrt{1 - x^2} \]Substitute back into (II) to obtain the final answer:
\[ \boxed{\dfrac{dy}{dx} = \dfrac{d(\arcsin x)}{dx} = \dfrac{1}{\sqrt{1 - x^2}}} \]Note that the above result could have been obtained using the formula theorem, but here we have shown how to find the derivative of the inverse without remembering the formula.
Exercises
Practice Problems & Solutions
Find the derivative of the inverse of each function given below:
- \( f(x) = 3x - 4 \)
- \( g(x) = \arccos x \)
- \( h(x) = \arctan x \)
Answers to the Above Exercises
- \( (f^{-1})'(x) = \dfrac{1}{3} \)
- \( (g^{-1})'(x) = -\dfrac{1}{\sqrt{1 - x^2}} \)
- \( (h^{-1})'(x) = \dfrac{1}{\sqrt{1 + x^2}} \)