# Thread: the chain rule using trigonometric functions

1. ## the chain rule using trigonometric functions

Will someone please tell me if I am doing this problem correctly?

2. You are close.

The correct derivation is

$g'(\theta) = \frac{d}{dx}\left(\sec\frac{\theta}{2}\tan\frac{\t heta}{2}\right)$
$= \sec\frac{\theta}{2}\left(\frac{1}{2}\sec^2\frac{\ theta}{2}\right)+\left(\frac{1}{2}\sec\frac{\theta }{2}\tan\frac{\theta}{2}\right)\tan\frac{\theta}{2 }$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\tan^2\frac{\theta}{2}$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\left(\sec^2\frac{\theta}{2}-1\right)$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec ^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}$
$=\sec^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}.$

3. Originally Posted by robasc
Will someone please tell me if I am doing this problem correctly?

Almost, you need to use the product rule with the chain rule. Which is what it looks like you started doing, but something went wrong. Your first part is right. After the chain rule you should have: $\frac{1}{2} sec^3(\frac{\theta}{2}) + \frac{1}{2}tan(\frac{\theta}{2})sec(\frac{\theta}{ 2})$ Then use the chain rule again with $u = \frac{\theta}{2}$

4. oops! the plus sign was just a mistake but I still am not getting why I should have to use the chain rule further, is this to simplify a bit more?