Derivative of a trig function

**b521** · July 19th 2010, 06:51 PM

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

**adkinsjr** · July 19th 2010, 07:02 PM

Originally Posted by b521

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

You just need to apply the chain rule. The general formula for

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

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

$\frac{1}{2\sqrt{x}(1+x)}$

**b521** · July 20th 2010, 11:24 AM

Thanks! It makes much more sense to me now!

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