How to Completely Simplify the Derivative of $\sqrt{16-x^2}-4\cos^{-1}(x/4)$ I am trying to completely simplify the derivative of the following function: 

So far, I have gotten the answer:

Apparently this is not simplified enough.
Does anyone know how to simplify this further, if it is at all possible?
All help is appreciated.
 A: \begin{align}
& \frac{d}{dx} \left(\sqrt{16-x^2}-4\cos^{-1}(\frac{x}{4})\right) \\[10pt]
= {} &\frac{d}{dx} \left(\sqrt{16-x^2}\right)-\frac{d}{dx} \left(4\cos^{-1} \left(\frac{x}{4}\right)\right) \\[10pt]
= {} & \frac{1}{2\sqrt{16-x^2}}\cdot(-2x)+\frac{4}{\sqrt{1-\frac{x^2}{16}}} \cdot \left(\frac{-1}{4}\right) \\[10pt]
= {} & \frac{-x}{\sqrt{16-x^2}}-\frac{4}{\sqrt{16-x^2}} \\[10pt]
= {} & \frac{-(4+x)}{\sqrt{16-x^2}}=\frac{-(4+x)}{\sqrt{(x+4)(x-4)}}=\frac{-1}{\sqrt{(x-4)}}.
\end{align}
I think this is right, double check.
A: You seem to have treated $\cos^{-1}\left(\dfrac x 4\right)$ as if it were $\left( \cos\left(\dfrac x 4\right) \right)^{-1}$.  But in fact in conventional usage $\text{“}\cos^{-1}\text{''}$ means the arccosine function --- i.e. the inverse function.
We have
$$
\frac d {dx} \arccos x = \frac{-1}{\sqrt{1-x^2}}.
$$
Use that.
A: use the substitution x=4cos(z).
A: The fully simplified answer is (4-x)/(sqrt(16-x^2)).
Simplify the (4cos^-1(x/4))' and you get -4/(sqrt(16-x^2)).
Subtract that from 1/(sqrt(16-x^2)), and you get (4-x)/(sqrt(16-x^2)).
