derivatives of inverse functions and the chain rule


In class we considered finding the derivatives of inverse functions. The key points to this were:

To illustrate, let's start by following the books' derivation of the derivative of the natural log function -- that is, we're trying to find (d/dx)(ln(x)). We know that

eln(x) = x.

Let's write z = ln(x) (so we're now trying to find dz/dx). Then we can write the equation as

ez = x.

Now let's differentiate both sides. On the left-hand side of the equation we have a composition: with y = ez, and z = ln(x), we're finding dy/dx. This is

(d/dx)(ez) = (d/dx)(x), or
(dy/dz)(dz/dx) = 1.

In the second line we have just filled in the chain rule. Here dy/dz = ez, so

(ez)(dz/dx) = 1.

But z = ln(x), so this is the same as

(eln(x))(dz/dx = 1, or
x (dz/dx) = 1.

Dividing by x, we get the result we expect:

(dz/dx = 1/x, or
(d/dx)(ln(x)) = 1/x.

Whew! Does that all make sense? If so, we're all but set: this is no different than finding the derivatives of inverse functions. Let's do a specific example of finding the derivative of an inverse function.

The derivative of arccosine

(This is a problem we did in class.) How can we find the derivative of the arccosine? Let's set up an equation that involves arccosine and differentiate it. The obvious equation is

cos(arccos(x)) = x.

Differentiating both sides, we have

(d/dx)(cos(arccos(x))) = (d/dx)(x), or
(d/dx)(cos(arccos(x))) = 1.

To find the derivative of the left-hand side we need the chain rule. Letting z = arccos(x) (so that we're looking for dz/dx, the derivative of arccosine), we get

(d/dx)(cos(z))) = 1, so
-sin(z) (dz/dx) = 1.

Dividing by -sin(z), we get

right triangle with angle z, adjacent leg x and hypotenuse 1
figure 1: graph of triangle with leg with length x and hypotenuse 1.
dz/dx = -1/sin(z) = -1/sqrt(1 - x2).

Thus we have our desired result: (d/dx)(arccos(x) = -1/sqrt(1 - x2).

The first part of this should be clear. Where did the square root come from? Think about the triangle shown to the right. We started off by saying cos(z) = x. We know that in a triangle, the cosine of an angle is the length of the adjacent side over the hypotenuse. So, thinking of x as x/1, we have a triangle with an angle z with adjacent side x and hypotenuse 1.

What's sin(z)? Sine is opposite/hypotenuse, so we need the length of the opposite side. Using the Pythagorean Theorem, we have (x)2 + (opposite)2 = 12, or (opposite) = sqrt(1 - x2). Thus sin(z) = sin(arccos(x)) = sqrt(1 - x2)/1 = sqrt(1 - x2).

derivatives of arbitrary inverse functions

Ok, one more level of abstraction! Let's consider inverse functions. Suppose that we want to find the derivative of the inverse function of a function f(x). The inverse function is f-1(x), and, by definition, has the property that

f ( f-1(x) ) = x.

So, to find the derivative of f-1(x), let's take the derivative of both sides of this equation. We suspect a chain rule might come in to this, so let's let z = f-1(x) first.

f(z) = x, so
(d/dx)(f(z)) = (d/dx)(x), or
(d/dx)(f(z)) = 1.

Applying the chain rule, we have

(f '(z))(dz/dx) = 1, so (dz/dx) = 1/f '(z).

Plugging back in for z, we get

(d/dx)(f-1(x) = 1/f '(f-1(x)).

What's the key point of all of this?

(Which is, of course, what we started with.)


derivatives of inverse functions
Last modified: Tue Oct 25 14:52:21 EDT 2005
Comments to:glarose(at)umich(dot)edu
©2005 Gavin LaRose, UM Math Dept.