<s>[INST] 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?
 [/INST] 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 $y'$   
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