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?
 A: 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.
A: It cries: "I am $\sqrt{x+4\sin x}$ up to a multiplicative constant!"
A: Troubled by the quotient rule? Then it might be helpful to rewrite your question:
$$\frac{\mathrm d}{\mathrm dx}\left(\frac{1 + 4\cos x}{2\sqrt{x+ 4\sin x}}\right) = \frac{1}{2} \left(\frac{\mathrm d}{\mathrm dx}(1 + 4\cos x)(x + 4\sin x)^{-1/2}\right)$$
Now, just use the product rule accompanied by the chain rule.
Try it out!

Here's how I would proceed:

Let $u = x + 4\sin x \implies u' = 1 + 4\cos x$
Substituting, your question now becomes:
$$\frac{1}{2}\left(\frac{\mathrm d}{\mathrm dx} (u') \cdot (u^{-1/2})\right)\\
= \frac{1}{2}\left(u''\cdot u^{-1/2}\ +\  u'\cdot \Big(-\frac{1}{2}u^{-3/2}\cdot u'\Big)\right)\\
= \frac{u''}{2\cdot u^{1/2}} - \frac{u'}{4\cdot u^{3/2}}\\
= \frac{2u\cdot u'' \ - \ u'}{4\cdot u^{3/2}}$$
where $u'' = \frac{\mathrm d}{\mathrm dx}\left( 1 + 4\cos x\right) =  - 4\sin x$ 
Now, you would get the same result if you had directly applied the quotient rule (which is a special case of the product rule). You can infact derive the quotient rule by applying the product rule to $\frac{\mathrm d}{\mathrm dx} f(x)\cdot\Big(g(x)\Big)^{-1         }$ . I leave this as an exercise to you but in order to convince yourself further, redo the question directly using the quotient rule       :     
$$ \frac{ \mathrm d}{\mathrm dx} \left(\frac{u'}{2\sqrt{u}}\right) = \frac{1}{2}\frac{\mathrm d}{\mathrm dx}\left(\frac{u'}{u^{1/2}}\right) = \frac{1}{2}\left(\frac{u''\sqrt u \ - \ u'\frac{1}{2 \sqrt{u}}}{u}\right) \equiv \frac{2u\cdot u'' \ - \ u'}{4\cdot u^{3/2}}$$ 
Alternatively, set $ y = \frac{u'}{2\sqrt u} \implies 2y\sqrt u = u'$ and then differentiate and solve to get $y'$   
A: Do it in order
$$\frac{\frac{d}{dx}(1+4\cos x)2\sqrt{x+4\sin x} -\frac{d}{dx}(2\sqrt{x+4\sin x})(1+4\cos x)}{4(x+4\sin x)}.$$
Solve the derivatives one by one and then replace in the big fraction. Remember to use the chain rule!
A: 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. 
