$\displaystyle
f(x) = \frac{cot x}{(1 + csc x)}
$
Find $\displaystyle f'(x)$
You know how to find the derivatives of $\displaystyle cot(x)$ and $\displaystyle csc(x)$ don't you? It's pretty easy if you know the derivatives of sine and cosine, so I won't go through it, I'll just write them down:
$\displaystyle \frac{d}{dx}cot(x)=-csc^2(x)$
$\displaystyle \frac{d}{dx}csc(x)=-csc(x)cot(x)$
So to find $\displaystyle f'(x)$, just apply the quotient rule.
$\displaystyle f'(x)=\frac{[1+csc(x)][-csc^2(x)]-cot(x)[-csc(x)cot(x)]}{[1+csc(x)]^2}$
$\displaystyle =\frac{[1-csc^2(x)]+cot^2(x)csc(x)}{[1+csc(x)]^2}$
EDIT: The last line is WRONG. I didn't use the distributive property on the first term. When I mulitplied $\displaystyle 1+csc(x)$ by $\displaystyle -csc^2(x)$ I should have obtained $\displaystyle -csc^2(x)-csc^3(x)$.
Ohh... I think I understand what you did :33
But I have the answer to the question && it doesn't match up.
The problem is that I can't get from the quotient rule step to the answer.
Sorry for complicating it so much >_<
The Answer:
$\displaystyle \frac{-csc x}{1+cscx}$