Originally Posted by

**ebonyscythe** Okay, well I tried the problem and I'm still having a hard time reasoning anything out. I've got the 2nd fund theorem of calc in front of me telling me that:

If function **S** is continuous on [a,b], then the function *D* is defined by

$\displaystyle D(x)=\int_{a}^{x}S(t)dt$ where $\displaystyle a\le x\le b$

and is continuous on [a,b], differentiable on (a,b) and

$\displaystyle D ' (x)=S(x)$

So, would that mean, in my problem above, that $\displaystyle f'(x)=\frac{1}{\sqrt{1+g(x)^3}}*g'(x)$?