http://i25.tinypic.com/20frvpj.png

I'm not sure how to solve this type of problem. Any help would be appreciated. Thanks.

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- Jul 19th 2010, 06:51 PMb521Derivative of a trig function
http://i25.tinypic.com/20frvpj.png

I'm not sure how to solve this type of problem. Any help would be appreciated. Thanks. - Jul 19th 2010, 07:02 PMadkinsjr
You just need to apply the chain rule. The general formula for

$\displaystyle \frac{d}{dx}Arctan(u)=\frac{u'}{1+u^2}$

Where $\displaystyle u$ is a function of x. You're basically applying the chain rule here. Just remember that the derivative of $\displaystyle u=\sqrt{x}$ is $\displaystyle u'=\frac{1}{2\sqrt{x}}$, and $\displaystyle u^2=x$. Once you put it all together, you should obtain:

$\displaystyle \frac{1}{2\sqrt{x}(1+x)}$ - Jul 20th 2010, 11:24 AMb521
Thanks! It makes much more sense to me now!