the chain rule using trigonometric functions

• Mar 27th 2009, 05:30 PM
robasc
the chain rule using trigonometric functions
Will someone please tell me if I am doing this problem correctly?

http://i244.photobucket.com/albums/g...g?t=1238203802
• Mar 27th 2009, 06:26 PM
Scott H
You are close.

The correct derivation is

$g'(\theta) = \frac{d}{dx}\left(\sec\frac{\theta}{2}\tan\frac{\t heta}{2}\right)$
$= \sec\frac{\theta}{2}\left(\frac{1}{2}\sec^2\frac{\ theta}{2}\right)+\left(\frac{1}{2}\sec\frac{\theta }{2}\tan\frac{\theta}{2}\right)\tan\frac{\theta}{2 }$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\tan^2\frac{\theta}{2}$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\left(\sec^2\frac{\theta}{2}-1\right)$
$=\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec ^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}$
$=\sec^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}.$
• Mar 27th 2009, 06:30 PM
mollymcf2009
Quote:

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

http://i244.photobucket.com/albums/g...g?t=1238203802

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 $u = \frac{\theta}{2}$
• Mar 27th 2009, 07:51 PM
robasc
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?