Derivative of trigonometric functions 
Find the derivative of $$\dfrac{5\sin x}{1-\cos x}.$$

I tried to do this myself by applying the product rule and got $$\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin x}{(1-\cos x)^2}.$$ I checked this with Wolfram Alpha but it was wrong. According to Wolfram Alpha, the derivative is $$\dfrac{5}{\cos(x) -1}.$$ Am i missing some simplification here?
 A: Use the quotient rule $\dfrac{f'g-g'f}{g^2}$. where $f(x) = 5\sin(x)$ and $g=1-\cos(x)$. You should obtain $\dfrac{-5}{(1-\cos(x))}$ which is indeed equal to wolfram alpha when you factor out the negative sign in the denominator.
A: The second term in the product rule seems to be missing a component due to the chain rule; the correct term should be
$$-5 \sin{x} \left(1 - \cos{x}\right)^{-2} \cdot \frac{d}{dx} \left(1 - \cos{x}\right) = -\frac{5 \sin^2{x}}{(1 - \cos{x})^2}$$
A: This is how I would simplify it:
if $$f(x)=g(x)h(x)$$
then $$f'(x) = h'(x)g(x)+g'(x)h(x)$$
so$$\dfrac{d}{dx}(\dfrac{5\sin x}{1-\cos x})$$
$$={\dfrac{d}{dx}(5\sin x)}(1-\cos x)^{-1}$$
$$=({\dfrac{d}{dx}(5\sin x)})*(1-\cos x)^{-1}+{\dfrac{d}{dx}(1-\cos x)^{-1}}*(5\sin x)$$
$$= (5\cos x)(1-\cos x)^{-1}+(-1)(1-\cos x)^{-2}(\sin x)(5\sin x)$$
$$=\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin ^2x}{(1-\cos x)^2}$$
Now we can multiply $\dfrac{5\cos x}{1-\cos x}$ by $\dfrac{1-\cos x}{1-\cos x}$
$$\dfrac{5\cos x}{1-\cos x} = \dfrac{5\cos x}{1-\cos x}*\dfrac{1-\cos x}{1-\cos x}$$
$$=\dfrac{5\cos x-5\cos ^2 x}{(1-\cos x)^2}$$
Now we can add the two fractions together
$$\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin ^2x}{(1-\cos x)^2}=\dfrac{5\cos x-5\cos ^2 x}{(1-\cos x)^2}-\dfrac{5\sin ^2x}{(1-\cos x)^2}$$
$$ = \dfrac{5\cos x-5\cos ^2 x -5\sin ^2x}{(1-\cos x)^2}$$
