# derivatives of inverse functions and the chain rule

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

*e*^{ln(x)} = *x*.

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

*e*^{z} = *x*.

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

(*d*/*dx*)(*e*^{z})
= (*d*/*dx*)(*x*), or

(*dy*/*dz*)(*dz*/*dx*) = 1.

In the second line we have just filled in the chain rule. Here
*dy*/*dz* = *e*^{z}, so

(*e*^{z})(*dz*/*dx*) = 1.

But *z* = ln(*x*), so this is the same as

(*e*^{ln(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 - *x*^{2}).

Thus we have our desired result:
(*d*/*dx*)(arccos(*x*) =
-1/sqrt(1 - *x*^{2}).

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} =
1^{2}, or
(opposite) = sqrt(1 - *x*^{2}).
Thus sin(*z*) = sin(arccos(*x*)) =
sqrt(1 - *x*^{2})/1 =
sqrt(1 - *x*^{2}).

## 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.)*

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.