Linear Inverse Functions
On the previous page we saw that if f(x)=3x + 1, then f has an inverse function given by f -1(x)=(x-1)/3. Both f and f -1 are linear funcitons.

An interesting thing to notice is that the slopes of the graphs of f and f -1 are multiplicative inverses of each other: The slope of the graph of f is 3 and the slope of the graph of f -1 is 1/3. This is a general feature of inverse functions. When you reflect across y=x, you take the reciprocal of the slope. This geometric observation goes a long way toward explaining the general differentiation formula for inverse functions.

Suppose that we are given a function f with inverse function f -1. Using a little geometry, we can compute the derivative Dx(f -1(x)) in terms of f.

The graph of a differentiable function f and its inverse are shown below. A point (x,y) has been selected on the graph of f -1. We have that f -1(x)=y.


The point (y,x) is on the graph of f, which means that f(y)=x. We denote the tangent line to the graph f at (y,x) by M. The slope of M is equal to the value of the derivative of f at y.

slope of M = f '(y) = f '(f -1(x))

Assuming that f -1 is differentiable at x, the slope of the tangent line will be given by the value of the derivative of f -1 at x. We denote the tangent line to f -1 at (x,y) by L. We want to compute the slope of L.

slope of L = Dx(f -1(x))

Since the graph of f -1 is obtained from the graph of f by reflection across y=x, the same is true for the tangent lines. L is obtained from M by reflection across the line y=x. This implies that the slopes of L and M are reciprocals.

Dx(f -1(x)) = slope of L = 1/(slope of M) = 1/f '(f -1(x))

Here is the actual result.


Theorem. Suppose that f has an inverse function f -1. If f is differentiable at f -1(x) and f '(f -1(x)) is not equal to zero, then f -1 is differentiable at x and the following differentiation formula holds.

Derivative of the Inverse Function

-1(x) =

f '(f -1(x))

We shall not discuss the proof of this theorem here, other than to say that the proof is relatively difficult. A first step in the proof is to show that the inverse of a continuous function is continuous; this proof in turn requires an application of the Intermediate Value Theorem (see Stage 4) and actually reveals a fairly deep and subtle property of the real numbers.


We have been more careful than usual in our statement of the differentiability result for inverse functions. You should notice that the differentiation formula for the inverse function involves division by f '(f -1(x)). We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity.

Example. The graphs of the cubing function f(x)=x3 and its inverse (the cube root function) are shown below.

y = x3
y = x1/3

Notice that f '(x)=3x2 and so f '(0)=0. The cubing function has a horizontal tangent line at the origin. Taking cube roots we find that f -1(0)=0 and so f '(f -1(0))=0. The differentiation formula for f -1 can not be applied to the inverse of the cubing function at 0 since we can not divide by zero. This failure shows up graphically in the fact that the graph of the cube root function has a vertical tangent line (slope undefined) at the origin.


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,infty) 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,infty). (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.
  • (x1/3)3 = x

  • 3(x1/3)2 Dx(x1/3) = 1

  • Dx(x1/3) = 1/(3x2/3) = (1/3)x-2/3

  • /Stage6/Lesson/inverseDeriv.html


    © CalculusQuestTM
    Version 1
    All rights reserved---1996
    William A. Bogley
    Robby Robson