How to get the derivative How do I find the derivative of $\displaystyle \frac{1+4\cos x}{2 \sqrt{x+4\sin x}}?$
So I got this as my answer: 
$$\frac{(4 \cos x+1)^2}{4\sqrt{4 \sin x+x}}-2 \sin x \sqrt{4\sin x+x}$$
does this look correct?

Looks like you will need to use the quotient rule and the chain rule to solve this. The quotient rule tells us that $$\left[\frac{1+4\cos x}{2 \sqrt{x+4\sin x}}\right]' = \frac{[1+4\cos(x)]'2 \sqrt{x+4\sin x}-\left(1+4\cos(x)\right)\left[2 \sqrt{x+4\sin(x)}\right]'}{\left(2 \sqrt{x+4\sin(x)}\right)^2}$$ and to evaluate $\left[2 \sqrt{x+4\sin(x)}\right]'$ you need the chain rule. You should get $$\left[2 \sqrt{x+4\sin(x)}\right]' = \frac{2}{2\sqrt{x+4\sin(x)}} \cdot \left(1+4\cos(x)\right) \\ =  \frac{1+4\cos(x)}{\sqrt{x+4\sin(x)}}$$ Now you can plug this expression into the one above to get $$ \frac{[1+4\cos(x)]'2 \sqrt{x+4\sin x}-\left(1+4\cos(x)\right) \frac{1+4\cos(x)}{\sqrt{x+4\sin(x)}}}{\left(2 \sqrt{x+4\sin(x)}\right)^2}$$ There is still a tremendous amount of simplification to be done, but it's mostly algebra. 
