subject: Inverse Functions And Their Derivatives [print this page] Let us consider two functions p and qLet us consider two functions p and q. If p (q(x)) = x and q (p(x)) = x then q is said to be the inverse of p and p is said to be the inverse of q. Then the functions p and q are referred to as the inverse functions. In other words if f is a function such that f: A -> B then its inverse function is given by f-1 (f(x)) = x.
How to Find Inverse Functions and their Derivatives
In the case inverse functions the statement y = f(x) and x = f-1(y) are equal, then the derivatives of the functions are given by
(dx / dy) * (dy / dx) = 1
The derivatives are the straight result of the chain rule which is given by
(dx / dy) * (dy / dx) = dx / dx
The derivative of x with respect to that of x is 1.
Ex 1: Find the derivative of y = x^2 which has the inverse of x = y
Given: y = x^2 x = y
dy / dx = 2x dx / dy = 1 / 2y
(dy / dx) * (dx / dy) = 2x * 1 / 2y
= 2x / 2x
= 1.
Ex 2: Find the derivatives of y = ex and its inverse x = ln (y)
Given: y = ex x = ln (y)
dy / dx = ex dx / dy = 1 / y
(dy / dx) * (dx / dy) = ex * 1 / y
= ex / ex
=1
Derivative of the Inverse Function:
Suppose we consider that f has an inverse function f-1. If f is said to be differentiable at f-1(x) and if f( f-1(x)) is not equal to zero, then f-1 is differentiable at x and the differentiation formula gives the derivatives of the inverse function.
d/dx (f-1(x)) = 1 / f (f-1(x))
Higher derivatives:
The higher derivatives of the inverse functions are given by
The chain rule makes it easy to differentiate inverse functions.
Example. The square root function is the inverse of the squaring function f(x)=x2. We must restrict the domain of the squaring function to [0,) in order to pass the horizontal line test. The differentiability theorem for inverse functions guarantees that the square root function is differentiable at x whenever f '(x)=2x is not equal to zero. This means that the square root function is differentiable on the open interval (0,). (We have already verified this using the limit definition of derivative. See Example 3 on the page "Differentiability: More Examples.")
Using the chain rule, we recalculate the derivative of the square function as follows.
The square of the square root of x is x.
Differentiate both sides. Use the chain rule on the left side.
Solve for the derivative of sqrt(x)!!
(sqrt(x))^2 = x
2sqrt(x) Dx(sqrt(x)) = 1
Dx(sqrt(x)) = 1/(2sqrt(x))
Example. The same trick works for the cube root function.
