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

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

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- Mar 27th 2009, 05:30 PMrobascthe 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 PMScott H
You are close.

The correct derivation is

$\displaystyle g'(\theta) = \frac{d}{dx}\left(\sec\frac{\theta}{2}\tan\frac{\t heta}{2}\right)$

$\displaystyle = \sec\frac{\theta}{2}\left(\frac{1}{2}\sec^2\frac{\ theta}{2}\right)+\left(\frac{1}{2}\sec\frac{\theta }{2}\tan\frac{\theta}{2}\right)\tan\frac{\theta}{2 }$

$\displaystyle =\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\tan^2\frac{\theta}{2}$

$\displaystyle =\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec \frac{\theta}{2}\left(\sec^2\frac{\theta}{2}-1\right)$

$\displaystyle =\frac{1}{2}\sec^3\frac{\theta}{2}+\frac{1}{2}\sec ^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}$

$\displaystyle =\sec^3\frac{\theta}{2}-\frac{1}{2}\sec\frac{\theta}{2}.$ - Mar 27th 2009, 06:30 PMmollymcf2009
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: $\displaystyle \frac{1}{2} sec^3(\frac{\theta}{2}) + \frac{1}{2}tan(\frac{\theta}{2})sec(\frac{\theta}{ 2})$ Then use the chain rule again with $\displaystyle u = \frac{\theta}{2}$

- Mar 27th 2009, 07:51 PMrobasc
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?