# Derivative of a trig function

• July 19th 2010, 06:51 PM
b521
Derivative 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.
• July 19th 2010, 07:02 PM
Quote:

Originally Posted by b521
http://i25.tinypic.com/20frvpj.png

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:

$\frac{1}{2\sqrt{x}(1+x)}$
• July 20th 2010, 11:24 AM
b521
Thanks! It makes much more sense to me now!