How can I study in a straightforward way this function with complicated derivatives? The study of this function 
$$f: \frac{\sin x}{e^{\frac{1}{1+\cos x}}}$$
seems quite awkward to me.
I could easily see that it is periodic ($T=2\pi$), that it is positive in $[0,\pi[$, odd, and that for $x\to\pi^-$, $f \to 0$. However, the expression of the first derivative seems extremely complicated, which makes it difficult to verify the monotonicity and the local maxima. The second derivative is even more complicated. How can I complete the analysis of the function $f$ (possibly in a clean and straightforward way)?
 A: Another way is
$$
f = \frac{\sin x}{\mathrm{e}^{g(x)}} = \mathcal{Im}\left(\mathrm{e}^{ix - g}\right)
$$
where $g(x) = \frac{1}{1+\cos x}$
thus
$$
f' = \mathcal{Im}\left[(i -g')\mathrm{e}^{ix - g}\right] = \mathrm{e}^{- g}\mathcal{Im}\left[(i -g')\mathrm{e}^{ix}\right] \\
=\mathrm{e}^{- g}\mathcal{Im}\left[(i -g')(\cos x + i \sin x)\right] \\
= \mathrm{e}^{- g}\left(\cos x - g'\sin x\right)
$$
for other derivatives 
$$
f'' = -g'f + \mathrm{e}^{- g}\left(-\sin x - g''\sin x -g'\cos x\right)
$$
A: The derivative is not so complex. May be, you could use logarithmic differentiation $$f=\frac{\sin x}{e^{\frac{1}{1+\cos x}}}$$ $$\log(f)=\log(\sin(x))-\frac{1}{1+\cos x}$$ $$\frac{f'}{f}=\cot (x)-\frac{\sin (x)}{(\cos (x)+1)^2}$$ To solve $f'=0$, use the tangent half-angle substitution $t=\tan(\frac x2)$. This would give you a very simple equation to solve for $t$.
A: As the OP notes, the function is periodic with period $2\pi$.  It's also an odd function, so it suffices to study it in the interval $[-\pi,\pi]$.  It has zeroes at $x=0$ and $\pm\pi$ and is otherwise positive in $(0,\pi)$ and negative in $(-\pi,0)$.  The exponential term's approach to infinity as $x\to\pm\pi$ means the function and all its derivatives are $0$ there.  In particular, the function flattens out at $\pm\pi$.
Putting all this together, we expect $f$ to have a single maximum between $0$ and $\pi$ and a single inflection point between the maximum and $\pi$.  (It could, of course, have a multitude of max's, min's, and inflection points; I'm just saying what we expect.)  Let's see.
Abbreviating $\cos x$ and $\sin x$ to $c$ and $s$ for notational convenience and writing $f=se^{-1/(1+c)}$, we have
$$f'=\left(s'-s\left(1\over1+c\right)' \right)e^{-1/(1+c)}=\left(c-{s^2\over(1+c)^2} \right)e^{-1/(1+c)}$$
Using the trig identity $s^2=1-c^2$, the final rational expression cleans up to give
$$f'=\left(c^2+2c-1\over c+1\right)e^{-1/(1+c)}$$
The local max/min's occur when $c^2+2c-1=0$, which is to say, when $c=\sqrt2-1$ (remembering that $-1\le c\le1$, which dismisses the other root, $-\sqrt2-1$, of the quadratic). Taking the inverse, you get $x\approx\pm1.1437$ rad, or $\pm65.53^\circ$.
So far, so good.
For the second derivative, it makes sense to rewrite
$$f'=\left(c+1-{2\over c+1}\right)e^{-1/(1+c)}=\left({1\over u}-2u \right)e^{-u}$$
where $u=1/(1+c)$.  Thus
$$f''=\left(-{1\over u^2}-2-{1\over u}+2u \right)u'e^{-u}=\left(2u^3-2u^2-u-1\over u^2 \right)u'e^{-u}$$
If you pick through all this carefully, you find that the second derivative is $0$ at $x=0$, $\pm\pi$, and when $c=\cos x=\sqrt6-3$.  (The cubic factors as $(u+1)(2u^2-4u-1)$, but $u=-1$ and one of the quadratic roots violate the condition $-1\le c\le 1$.)  So the remaining inflection point is at $x\approx2.1538$ rad, or $123.4^\circ$.
