1st derivatives of $f(\alpha) = \frac{\sin(2\alpha)}{\sin(\alpha+1)}$ Could someone help me out with the following?
I have to get a maximum using the derivative
$$f(\alpha) = \frac{\sin(2\alpha)}{\sin(\alpha+1)}$$
$$f(\alpha) = \sin(2\alpha) \cdot (\sin(\alpha+1))^{-1}$$
$$f'(\alpha) = \sin(2\alpha) \cdot ((\sin(\alpha+1))^{-1})' + (\sin(2\alpha))' \cdot (\sin(\alpha+1))^{-1}$$
$$f'(\alpha) = \sin(2\alpha) \cdot \color{red}{\cdots} + 2 \cos(2\alpha) \cdot (\sin(\alpha+1))^{-1}$$
I can't get any furher then this
 A: Use the quotient rule:
$$
\left(\frac{f(x)}{g(x)} \right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}
$$
here we have $f(x)=\sin(2x)$ and $g(x)=\sin(x+1)$. So we have
$$
\frac{2\cos(2x)\sin(x+1)-\sin(2x)\cos(x+1)}{\sin^2(x+1)}
$$
If you are looking for a maximum, set this equal to zero and solve.
EDIT. As it has been suggested, the intended problem was 
$$
\left(\frac{\sin(2x)}{\sin(x)+1} \right)'
$$
(always a good idea to include parenthesis around function arguments where confusion could arise!)
So that the derivative of the denominator is $\left(\sin(x)+1\right)'=\cos(x)+0=\cos x$. That would give the solution:
$$
\frac{2\cos(2x)\left(\sin(x)+1\right)-\sin(2x)\cos(x)}{\left(\sin(x)+1\right)^2}
$$
which matches their solution.
A: As mathematics2x2life has pointed out, the quotient rule can be used.  However, I will address directly the question of what to do with
$$
\frac{d}{d\alpha} (\sin(\alpha+1))^{-1}.
$$
You can say it's $\dfrac{d}{d\alpha} u^{-1}$, so that it's
$$
\frac{d}{d\alpha} u^{-1} = (-1)u^{-2}\cdot\frac{du}{d\alpha} = -u^{-2}\cdot\frac{d}{d\alpha} \sin(\alpha+1) = \frac{-1}{u^2}\cdot\cos(\alpha+1). 
$$
After that you would multiply by $\dfrac{d}{d\alpha}(\alpha+1)$, but you'd just be multiplying by $1$, so that doesn't change anything.  Then you have
$$
\frac{-1}{(\sin(\alpha+1))^2}\cdot\cos(\alpha+1).
$$
