# derivatives of inverse functions and the chain rule

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

• If we have an equation that involves a function (in particular, an inverse function), we can differentiate both sides of the equation, and
• If we differentiate a composition we have to use the chain rule, which can introduce factors of the (unknown) derivative we're trying to find.

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

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?

• We can differentiate both sides of an equation, and
• Whenever we have a function composed with another function we need the chain rule.

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

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derivatives of inverse functions