# Chain rule review

Review your knowledge of the Chain rule for derivatives, and use it to solve problems.

## What is the Chain rule?

The Chain rule tells us how to differentiate expressions that are the composition of two other, more basic, expressions:
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
Basically, you compose the derivative of f with g, and multiply it by the derivative of g.
Want to learn more about the Chain rule? Check out this video.

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

### Example 1

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.
\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}

### Check your understanding

Problem 1
start fraction, d, divided by, d, x, end fraction, left parenthesis, cosine, left parenthesis, e, start superscript, x, end superscript, right parenthesis, right parenthesis, equals, question mark
Please choose from one of the following options.

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

### Example 2

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}

### Check your understanding

Problem 1
xf, left parenthesis, x, right parenthesish, left parenthesis, x, right parenthesisf, prime, left parenthesis, x, right parenthesish, prime, left parenthesis, x, right parenthesis
minus, 19minus, 1minus, 5minus, 6
23minus, 116
G, left parenthesis, x, right parenthesis, equals, f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis
G, prime, left parenthesis, 2, right parenthesis, equals

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