## Differentiation As An Operator

The sum, product, quotient, and chain rules make it possible to differentiate many functions. However, there are some more exotic functions which cannot be differentiated using these tools alone. For example, what is the derivative of or or ?

The derivative should be interpreted as a rate of change, but what about the act of differentiation? Differentiation is an *operator:* it takes in a function and gives out another function. Other examples of operators include the logarithm, exponentiation, and integration. These (and other) operators can be applied to an entire equation to transform a hard problem to an easy problem and (once the solution is found) back again. This idea will allow us to compute the derivatives of the exotic functions above, and more.

## Logarithmic differentiation

A common combination is called *logarithmic differentiation,* which consists of applying the logarithm operator followed by the differentiation operator. It is best demonstrated by example.

**Example**

Find the derivative of using logarthmic differentiation.

We already know the derivative of , but suppose we did not. Let . Then applying the logarithm operator to this equation gives

Now applying the differentiation operator to the equation (and remembering the chain rule) gives

so , and since from our original definition, we have

as expected.

**Example**

Use logarithmic differentiation to show that .

Similarly to the above example, let . Then taking the logarithm gives

Differentiating gives

And so

as desired.

**Example**

Find the derivative of .

Let . Taking the logarithm gives

Differentiating (using the product rule on the right) gives that

Next, factoring and solving for gives

**Example**

Find the derivative of .

Let . Applying the logarithm gives , and differentiating gives

Factoring and solving for shows

## Other operators

There are other operators which can be used prior to differentiation. Consider the following examples.

**Example**

Compute the derivative of by using exponentiation followed by differentiation.

Letting , we exponentiate the equation to find

Now, differentiating gives

and solving for gives

as desired.

Note that is the inverse of , and so when we exponentiated the equation, it could be thought of as applying the inverse of . This same method works for many other inverse functions. In particular, applying a trigonometric function can be thought of as an operator as well. This method can be used to find the derivatives of the various inverse trigonometric functions.

**Example**

Find the derivative of .

Letting , take the sine of the equation to find

Now, differentiating gives

Thus, we find that

Now, a little bit of trigonometry helps rewrite in terms of . Our original equation had . That means that is the angle such that . Since sine is the opposite over the hypotenuse, we can express this relationship with the following right triangle:

where the adjacent leg comes from the Pythagorean theorem. It follows that , and so we find

**Example**

Find the derivative of .

Let . Applying to the equation gives

and differentiating gives

Therefore,

We can now do similar right triangle trig as in the previous example. Or we can recall that by the Pythagorean identity for tangent and secant, we have

Making this substitution gives

## Operators in other contexts

Besides being useful in computing derivatives of exotic functions, operators (especially the logarithm) can also be useful in computing limits. The method is similar to the above method for derivatives.

**Example**

Show that

This is a common limit which will come up again in the course.

Let be the function given by

Take the logarithm of both sides, and use the power property of logarithms to see

Now, taking the limit of both sides gives

Now, since , we have that is small, and so we can use our Taylor series for , which gives us that

Recall from our limit rules that the order of the logarithm and the limit can be switched since the logarithm is a continuous function. Thus

So, exponentiating both sides, we have that

as desired.

## Infinite Power Tower

Consider the infinite power tower

That is, raised to the raised to the etc. This is certainly an unusual function. A better way to define this function is implicitly:

To see that this makes intuitive sense, note that the exponent of the first in the infinite tower is *itself* an infinite tower, so replacing the exponent of with is sensible.

Use logarithmic differentiation to find the derivative of this function (that is, ).

First, taking the logarithm gives

Now, differentiating the equation (implicitly) gives

Factoring and solving for gives

This shows that although a function may be difficult to understand, it can nevertheless be fairly easy to find its derivative.

It turns out that this function is well-defined and differentiable on

One can check that the pairs all satisfy the above implicit equation.

## EXERCISES

- Find the derivative of
- Find the derivative of
- Compute
- Compute
- Compute
- Compute
- Let

- Let