Paul's Online Notes
Home / Calculus I / Derivatives / Chain Rule
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 3.9 : Chain Rule

26. Differentiate $$h\left( z \right) = {\tan ^4}\left( {{z^2} + 1} \right)$$ .

Show All Steps Hide All Steps

Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the derivative.
Start Solution

This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative process to make each use clear. Also, recall that,

${\tan ^4}\left( x \right) = {\left[ {\tan \left( x \right)} \right]^4}$

Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.

$h'\left( z \right) = 4{\tan ^3}\left( {{z^2} + 1} \right)\frac{d}{{dz}}\left[ {\tan \left( {{z^2} + 1} \right)} \right]$ Show Step 2

As we can see the derivative from the previous step will also require the Chain Rule.

The derivative is then,

$h'\left( z \right) = 4{\tan ^3}\left( {{z^2} + 1} \right){\sec ^2}\left( {{z^2} + 1} \right)\left( {2z} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{8z{{\tan }^3}\left( {{z^2} + 1} \right){{\sec }^2}\left( {{z^2} + 1} \right)}}$