# Chain rule review

Review your knowledge of the Chain rule for derivatives, and use it to solve problems.
The chain rule says:
start fraction, d, divided by, d, x, end fraction, open bracket, f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, close bracket, equals, f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis
It tells us how to differentiate functions that are the composition of two more basic functions—an inner function g, left parenthesis, x, right parenthesis and an outer function f, left parenthesis, x, right parenthesis.

## What problems can I solve with the chain rule?

The chain rule only applies to composite functions—functions that can be written as start color greenD, f, left parenthesis, end color greenD, start color goldD, g, left parenthesis, x, right parenthesis, end color goldD, start color greenD, right parenthesis, end color greenD. So, naturally, a key skill in using the chain rule is being able to identify composite functions. If a function isn't composite, we can't use the chain rule.
For example, h, left parenthesis, x, right parenthesis, equals, left parenthesis, 5, minus, 6, x, right parenthesis, start superscript, 5, end superscript can be viewed as
where the inner and outer functions are:
\begin{aligned} \goldD{g(x)}&=\goldD{5-6x} &&\text{inner function} \\\\ \greenD{f(x)}&=\greenD{x^5}&&\text{outer function} \end{aligned}
Because we've got a composite function on our hands, we can differentiate using the chain rule. But before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown):
\begin{aligned} \maroonD{g'(x)}&=\maroonD{-6} \\\\ \blueD{f'(x)}&=\blueD{5x^4} \end{aligned}
Now let's apply the chain rule:
\begin{aligned} &\dfrac{d}{dx}\left[f\Bigl(g(x)\Bigr)\right] \\\\ =&\blueD{f'\Bigl(\goldD{g(x)}\Bigr)}\cdot\maroonD{g'(x)} \\\\ =&\blueD{5(\goldD{5-6x})^4} \cdot \maroonD{-6} \\\\ =&-30(5-6x)^4 \end{aligned}

## Common misconceptions

Here are three common student misconceptions when applying the chain rule:
Confusing products and compositions: Especially with transcendental functions (e.g., trigonometric and logarithmic functions), students often confuse products like natural log, left parenthesis, x, right parenthesis, sine, left parenthesis, x, right parenthesis with compositions like natural log, left parenthesis, sine, left parenthesis, x, right parenthesis, right parenthesis. The chain rule only applies to function compositions—we'd need to use the product rule for products.
Forgetting to multiply by the derivative of the inner function: In the example above, we multiplied by start color maroonD, g, prime, left parenthesis, x, right parenthesis, end color maroonD, but many students forget this part. That is, some students compute start color blueD, f, prime, left parenthesis, start color goldD, g, left parenthesis, x, right parenthesis, end color goldD, right parenthesis, end color blueD instead of start color blueD, f, prime, left parenthesis, start color goldD, g, left parenthesis, x, right parenthesis, end color goldD, right parenthesis, end color blueD, dot, start color maroonD, g, prime, left parenthesis, x, right parenthesis, end color maroonD.
Computing f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis: If you're not paying close attention, you might accidentally compute f, prime, left parenthesis, g, prime, left parenthesis, x, right parenthesis, right parenthesis instead of f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis. Notice how g, left parenthesis, x, right parenthesis, not g, prime, left parenthesis, x, right parenthesis, is the function that should be inside of f, prime, left parenthesis, x, right parenthesis.

Consider the differentiation of tangent, left parenthesis, x, start superscript, 2, end superscript, right parenthesis. Notice that if u, left parenthesis, x, right parenthesis, equals, x, start superscript, 2, end superscript and v, left parenthesis, x, right parenthesis, equals, tangent, left parenthesis, x, right parenthesis, then tangent, left parenthesis, x, start superscript, 2, end superscript, right parenthesis, equals, v, left parenthesis, u, left parenthesis, x, right parenthesis, right parenthesis. We're dealing with a composite function, so we can apply the chain rule.
\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\tan(x^2)\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=x^2\text{, }v(x)=\tan(x)} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{Chain rule}} \\\\ &=\sec^2(x^2)\cdot 2x \\\\ &=2x\sec^2(x^2) \end{aligned}

Problem 1
start fraction, d, divided by, d, x, end fraction, open bracket, square root of, cosine, left parenthesis, x, right parenthesis, end square root, close bracket, equals, space, question mark
Please choose from one of the following options.

Want to try more problems like this? Check out this exercise.

### Examples with tables

Suppose we are given this table of values:
xf, left parenthesis, x, right parenthesisg, left parenthesis, x, right parenthesisf, prime, left parenthesis, x, right parenthesisg, prime, left parenthesis, x, right parenthesis
minus, 25minus, 116
4minus, 4minus, 208
H, left parenthesis, x, right parenthesis is defined as f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, and we are asked to find H, prime, left parenthesis, 4, right parenthesis.
The chain rule tells us that H, prime, left parenthesis, x, right parenthesis is f, prime, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, g, prime, left parenthesis, x, right parenthesis. This means H, prime, left parenthesis, 4, right parenthesis is f, prime, left parenthesis, g, left parenthesis, 4, right parenthesis, right parenthesis, g, prime, left parenthesis, 4, right parenthesis. Now let's plug the values from the table in the expression:
\begin{aligned} H'(4)&=f'\Bigl(g(4)\Bigr)\cdot g'(4) \\\\ &=f'(-2)\cdot 8&&\gray{g(4)=-2\text{ , }g'(4)=8} \\\\ &=1\cdot 8&&\gray{f'(-2)=1} \\\\ &=8 \end{aligned}