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?

To avoid that nasty quotient rule
(and throwing away that "$2$"),
$\begin{array}\\
\left(\frac{1+4\cos x}{ \sqrt{x+4\sin x}}\right)'
&=\left((1+4\cos x)(x+4\sin x)^{-1/2}\right)'\\
&=(1+4\cos x)\left((x+4\sin x)^{-1/2}\right)'
+(1+4\cos x)'(x+4\sin x)^{-1/2}\\
&=(1+4\cos x)\left((-1/2)((x+4\sin x)'(x+4\sin x)^{-3/2}\right)
+(-4\sin x)(x+4\sin x)^{-1/2}\\
&=(1+4\cos x)\left((-1/2)((1+4\cos x)(x+4\sin x)^{-3/2}\right)
+(-4\sin x)(x+4\sin x)^{-1/2}\\
&=\frac{(-1/2)(1+4\cos x)^2
+(-4\sin x)(x+4\sin x)}{(x+4\sin x)^{3/2}}\\
&=-\frac{(1+4\cos x)^2
+(8\sin x)(x+4\sin x)}{2(x+4\sin x)^{3/2}}\\
&=-\frac{1+8\cos x+16\cos^2 x
+8x\sin x+32\sin^2 x}{2(x+4\sin x)^{3/2}}\\
&=-\frac{1+8\cos x+16
+8x\sin x+16\sin^2 x}{2(x+4\sin x)^{3/2}}\\
\end{array}
$
Or,
looking at 
Przemysław Scherwentke's
marvelous suggestion,
since
$(x+4\sin x)'
=1+4\cos x
$
and
$(f'f^{-1/2})'
=f'' f^{-1/2}+f'(-1/2)f'f^{-3/2}
=\dfrac{f'' f+(-1/2)f'^2}{f^{-3/2}}
$,
if
$f(x)
=x+4\sin x
$,
$f'(x)
=1+4\cos x
$
and
$f''(x)
=-4\sin x
$
we get
$\begin{array}\\
(f'f^{-1/2})'
&=f'' f^{-1/2}+f'(-1/2)f'f^{-3/2}\\
&=\dfrac{(-4\sin x) (x+4\sin x)+(-1/2)(1+4\cos x)^2}{(x+4\sin x)^{3/2}}\\
&=\dfrac{-4x\sin x-16\sin^2 x)-(1/2)(1+8\cos x+16\cos^2x}{(x+4\sin x)^{3/2}}\\
&=-\dfrac{8x\sin x+32\sin^2 x+1+8\cos x+16\cos^2x}{(x+4\sin x)^{3/2}}\\
\end{array}
$.
By a miracle,
the two results are the same.
