Can't get the right result for a derivative of a trigonometric function so I need to find the derivative of the following expression:
$$y=(-\csc x)(-\sin x)$$
This is what I have done so far by applying the product rule:
$$y'=\csc x\cot x(-\sin x) + (-\csc x)(-\cos x)=-\sin x\csc x\cot x + \csc x\cos x$$
$$y'=\csc x(-\sin x\cot x + \cos x)$$
Unfortunately, my textbook displays the result as:
$$y'=\cos x\cot^2x$$
Am I doing something wrong, or both results are equivalent? If they are equivalent, can someone show me step by step how to get to the textbook's result?
Thank you so much in advance!
 A: The textbook's answer of $y' = \cos x \cot^2 x$ is correct for the very similar-looking function $$y = -\csc x - \sin x$$ for which we have
$$
   y' = \cot x \csc x - \cos x = \cos x \cdot (\csc^2 x - 1) = \cos x \cot^2 x.
$$
So the source of the discrepancy is probably a mistake on someone's part:


*

*Either the authors of the textbook meant to ask the derivative of $-\csc x - \sin x$ instead of $(-\csc x)(-\sin x)$,

*Or the authors confused the two later, when writing solutions to the exercises,

*Or you misread the exercise and it does say $-\csc x - \sin x$.

A: Notice:
$$-\csc x \times -\sin x = \frac{1}{\sin x} \times \sin x =1$$
Hence $f'(x)=\frac{\mathrm{d}}{\mathrm{dx}}(1)=0$
A: Your result is correct as it is $0$ (wherever $\csc x$ is defined). The book’s is not. 
A: It might be easier to look at everything in terms of the two "main" functions, sine and cosine. It's easier to see whether answers are equivalent this way, and the (anti)derivatives are easier to remember as well. In that light, the textbook's answer is
$$y' = \cos(x) \cot^2(x) = \cos(x) \cdot \frac{\cos^2(x)}{\sin^2(x)} = \frac{\cos^3(x)}{\sin^2(x)}$$
As for your answer,
$$y' = \csc(x) (-\sin(x)\cot(x) + \cos(x)) = \frac{1}{\sin(x)} \left( -\sin(x) \cdot \frac{\cos(x)}{\sin(x)} + \cos(x) \right) = 0$$
so definitely not equivalent.

So with that out of the way, let's approach the problem in this same mindset: define $y$ by
$$y = (-\csc(x))(-\sin(x))$$
Then since $\csc(x) = 1/\sin(x)$, we immediately see $y = 1$:
$$y = \frac{-1}{\sin(x)} \cdot (-\sin(x)) = (-1)^2 = 1$$
Thus $y=1$ and $y'=0$ (at least wherever $\sin(x) \ne 0$ owing to the cosecant) so your answer is correct, in a sense, just not simplified very well. Your work seems fine, it's just roundabout and not simplified sufficiently.
As for the textbook's answer, I'm not sure how they came to that conclusion...
