Probably really easy, derivative of cotx/sinx

[dietcookie]

Homework Statement

Find the derivative of the function.

f(x) = $$\frac{cotx}{sinx}$$

Homework Equations

$$\frac{cotx}{sinx}$$ = $$\frac{cosx}{sin^{2}x}$$

The Attempt at a Solution

I have the solution guide, I think I'm just having a brain fart but this is what I have so far which is the last step before the final answer

$$\frac{-sin^{2}x - 2cos^{2}x}{sin^{3}x}$$

The next step, which is the answer according to the book is

$$\frac{-1-cos^{2}x}{sin^{3}x}$$

Am I just not seeing something on my last step? I think they got the -1 from using the pythagorean indentity but then why is the cosx still hanging around? Thanks.

 

[Borek]

You will feel ashamed enough to put a paper bag on the head.

$$-\sin^2x -2\cos^2x = -(\sin^2x+\cos^2x)-\cos^2x$$

 

[dietcookie]

You will feel ashamed enough to put a paper bag on the head.

$$-\sin^2x -2\cos^2x = -(\sin^2x+\cos^2x)-\cos^2x$$

Yep that's most definitely embarrassing! Thanks!