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Study Guide

How do we find the derivative of a function that's made of one function nested inside another, like

*e*^{sin x}

or

(ln *x*)^{42}?

The tool we need is called the **Chain Rule**. While the chain rule isn't always likeable at first, it grows on you. Once we've got the chain rule, we can also do something called **implicit differentiation**, which allows us to say some things about a function's derivative even if we aren't given a formula for that function. Clearly, our derivatives toolbox is growing.

### The Chain Rule

When dealing with a composition

*h*(*x*) =*f*(*g*(*x*)),the function

*f*is called the**outside function**and the function*g*is called the**inside function**.*f*is the outside function because it's written on the outside of the parentheses:*f*(*g*(*x*))*g*is the inside function because it's written on the inside of the parentheses:*f*(*g*(*x*)).

Written like this, the idea of outside and inside functions makes a lot of sense, but somehow it always gets more complicated when we start looking at actual functions.### "Outside" and "Inside" Functions

Another way to think of outside and inside functions in the case of the composition

*h*(*x*) =*f*(*g*(*x*))is to think about what we need to do in order to evaluate

*h*at a particular value of*x*. First we would need to find the output of*g*at that value of*x*. Then we would put that output into*f*. The inside function creates the input that we need for the outside function. Like the product and quotient rules, the outside function is the one we evaluate last.### Sample Problem

Look at the function

*h*(*x*) =*e*^{sin x}.This is a composition of the function

*e*^{{(□)}}and sin*x*. In order to evaluate this function at , the first thing we dois find the output of sin at this value of

*x*:The next thing we do is find

*e*raised to that power:For this composition, sin

*x*is the inside function because sin*x*is the*input*to*e*^{{(□)}}. Therefore*e*^{{(□)}}must be the outside function. We're writing*e*^{{(□)}}instead of

*e*^{x}to emphasize that we'll be raising

*e*to some other power besides*x*.The

*chain rule*states that(

*f*(*g*(*x*)))' =*f*' (*g*(*x*)) ·*g '*(*x*).If we state the chain rule with words instead of symbols, it says this: to find the derivative of the composition

*f*(*g*(*x*)),- identify the outside and inside functions

- find the derivative of the outside function and then use the original inside function as the input

- multiply by the derivative of the inside function.

Of course, we also want to simplify the answer into something reasonable.

Some people like thinking about the chain rule as

*f*'(*g*(*x*)) ·*g '*(*x*),while some prefer

(derivative of outside function evaluated at inside function)(derivative of inside function).

Some prefer Leibniz notation. Use the way that works best.

**Be Careful:**When using the chain rule, use the original inside function as input to the derivative of the outside function. Then remember to multiply by the derivative of the inside function.We can use the chain rule on functions that are nested 3 or more deep; we just need to write things down carefully.

- identify the outside and inside functions
### Re-Constructing the Quotient Rule

One of the nifty tricks we can do with the chain rule is reconstruct the quotient rule. To find the derivative of

first rewrite the function like this:

*h*(*x*) =*f*(*x*) · (*g*(*x*))^{-1}.Now that

*h*is written as a product, we can use the product rule to find its derivative:*h '*(*x*) =*f '*(*x*)(*g*(*x*))^{-1}+*f*(*x*) · ((*g*(*x*))^{-1})'.Now there are two big steps left. First we'll use the chain rule to find this derivative:

*h '*(*x*) =*f '*(*x*)(*g*(*x*))^{-1}+*f*(*x*) · ((*g*(*x*))^{-1})'.Then we'll simplify the formula we got using the product rule until it magically turns into the quotient rule.

Chain Rule:

To finish applying the product rule, we need to know

((

*g*(*x*))^{-1})'In other words, we need to know the derivative of the nested function

(

*g*(*x*))^{-1}Do our chain rule stuff. The outside function is

(□)

^{-1},and its derivative is

-(□)

^{-2}.The inside function is

*g*(*x*),and its derivative is

*g '*(*x*).Now we can use the chain rule:

((

*g*(*x*))^{-1})' = -(*g*(*x*))^{-2}·*g '*(*x*)Since (

*g*(*x*))^{-2}is the same thing as , we can rewrite this asReturning to the product rule,

We can do some great simplifying here. Since (

*g*(*x*))^{-1}is the same thing as , we can rewrite this asNow we can put the fractions over a common denominator and combine them:

Ta-daa! This is the quotient rule.

### Derivative of a

^{x}The functions

*e*and ln^{x}*x*are inverses of each other. The part we care about right now is that for any positive real number*a*,*e*^{ln a}=*a*.If we turn this equation around, we can write any positive real number

*a*as*e*^{ln a}.For example,

7 =

*e*^{ln 7}Therefore

7

^{x}is the same thing as

(

*e*^{ln 7})^{x}which by rules of exponents is equal to

*e*^{(ln 7)x}.We can find the derivative of

*h*(*x*) =*e*^{(ln 7)x}using the chain rule. The outside function is

*e*^{{□}},whose derivative is also

*e*^{{□}},and the inside function is

(ln 7)

*x*,whose derivative is the constant

(ln 7).

The chain rule says

*h '*(*x*) =*e*^{(ln 7)x}· (ln 7)Turning

*e*^{(ln 7)x}back into 7^{x}, we see that*h '*(*x*) = 7^{x}· (ln 7).This is where we find the rule for taking derivatives of exponential functions that are in other bases than

*e*.### Derivative of

*ln*xIf we didn't already know the derivative of ln

*x*, we could figure it out using the chain rule.We know that

*e*^{ln x}=*x*.Take the derivative of each side of this equation.

The derivative of

*x*is 1.

To find the derivative of*e*^{ln x}we need to use the chain rule. The outside function is*e*^{{□}}and its derivative is also*e*^{{□}.}

The inside function is ln*x*. Since we don't yet know the derivative of ln*x*(at least, we're pretending we don't) we'll write its derivative as (ln*x*)'.The chain rule says

(

*e*^{ln x})' =*e*^{ln x}· (ln*x*)'Since

*e*^{ln x}=*x*, we can simplify this to(

*e*^{ln x})' =*x*· (ln*x*)'Now return to the equation

*e*^{ln x}=*x*.The derivative of the right-hand side is 1, and the derivative of the left-hand side is

*x*· (*ln**x*)', therefore*x*· (ln*x*)' = 1.Dividing both sides by

*x*, we find### Derivatives of Inverse Trigonometric Functions

Finding derivatives of inverse trig functions will be similar to finding the derivative of ln

*x*. Our main tool is the chain rule. We also need some background information:- the fact that composing inverse functions gets us back where we started.

- the Pythagorean Theorem

- SOHCAHTOA

Here is a recap video of Pythagorean Theorem.

Here is a recap video on SOHCAHTOA.

- the fact that composing inverse functions gets us back where we started.
### The Chain Rule in Leibniz Notation

We stated the chain rule first in Lagrange notation. Since Leibniz notation lets us be a little more precise about what we're differentiating and what we're differentiating with respect to, we need to also be comfortable with the chain rule in Leibniz notation.

Suppose

*y*is a function of*x*:*y*=*g*(*x*)

and*z*is a function of*y*:

*z*=*f*(*y*)

Then*z*is a function of*x*:

*z*=*f*(*y*) =*f*(*g*(*x*))

Once again, we have an outside function and an inside function. The chain rule in Lagrange notation states that(

*f*(*g*(*x*))' =*f*' (*g*(*x*)) ·*g '*(*x*).In Leibniz notation, we would say

Since

- and (
*f*(*g*(*x*))' both mean "the derivative of*z*with respect to*x*,"

- and
*f*' (*g*(*x*)) =*f*' (*y*) both mean "the derivative of*z*with respect to*y*," and

- and
*g '*(*x*) both mean "the derivative of*y*with respect to*x*,"

the two statements of the chain rule do mean the same thing.

We can remember the chain rule in Leibniz notation because it looks like a nice fraction equation where the

*dy*terms cancel:This may or may not be what's actually going on, but it works for our purposes and it's a great memory aid.

There are three steps to apply the chain rule in this form:

- determine what
*y*is (this is the same step as determining the inside and outside functions)

- apply the chain rule formula

- put everything in terms of the correct variable (for example, writing
*y*in terms of*x*)

The chain rule will be especially useful when we discuss related rates, where there will be problems with three different variables that all depend on each other in funny ways.

We can also use the chain rule with different letters, as long as we put the letters in the correct places. If we have

=**y***g*(*x*) and*z*=*f*(*y*) =*f*(*g*(*x*)), the chain rule saysThe inside function is the one that "cancels out":

The innermost variable,

*x*, goes only in the denominators:and the outermost variable,

*z*, goes only in the numerators:If we switch the letters, we need to make sure they go in the appropriate places.

The important thing is that the inside function needs to be the term that cancels out:

Now that we know how to write the chain rule with different letters, we can use it to find derivatives.

- and (

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