<s>[INST] 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?
 [/INST] 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
$$\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin ^2x}{(1-\cos x)^2}=\dfrac{5\cos x-5\cos ^2 x}{(1-\cos x)^2}-\dfrac{5\sin ^2x}{(1-\cos x)^2}$$
$$ = \dfrac{5\cos x-5\cos ^2 x -5\sin ^2x}{(1-\cos x)^2}$$
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