Derivative of a product of trig functions: $3\sin(x)\cot(x)$ I am having a lot of trouble understanding on how to find the derivative of $3\sin x\cot x$.
I end up with $-3\cos x/\sin^2 x$
 A: While Michael Rybkin's answer is correct, it is worth noting that the expression $3\sin(x)\cot(x)$ is not in the simplest form, and begs the use of a different approach. Recalling that
$$\cot(x) = \frac{\cos(x)}{\sin(x)}$$
then
$$3\sin(x)\cot(x) = 3\sin(x) \frac{\cos(x)}{\sin(x)} = 3\cos(x)$$
Then the derivative is simply
$$\frac{d}{dx} 3\cos(x) = 3 \frac{d}{dx} \cos(x) = -3 \sin(x)$$

Of course, as noted in Michael's answer you can use the product rule,
$$\frac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$$
and as shown there you still obtain the same answer.
A: You need to use the product rule:
$$
(3\sin{x}\cot{x})'=\\
3(\sin{x})'\cot{x}+3\sin{x}(\cot{x})'=\\
3\cos{x}\cot{x}+3\sin{x}(-\csc^2{x})=\\
3\cos{x}\cot{x}-3\sin{x}\csc^2{x}=\\
3\cos{x}\frac{\cos{x}}{\sin{x}}-3\sin{x}\frac{1}{\sin^2{x}}=\\
3\frac{\cos^2{x}}{\sin{x}}-3\frac{1}{\sin{x}}=\\
3\frac{1}{{\sin{x}}}(\cos^2{x}-1)=\\
3\frac{1}{{\sin{x}}}(1-\sin^2{x}-1)=-3\sin{x}.
$$
A much simpler way to do this, as was mentioned in the comments section, is to notice that $3\sin{x}\cot{x}=3\sin{x}\frac{\cos{x}}{\sin{x}}=3\cos{x}$ which is trivial to differentiate:
$$
(3\cos{x})'=3(\cos{x})'=3(-\sin{x})=-3\sin{x}.
$$
