simplifying trig expression (I think)

Hello, infraRed!

Quote:

$\text{Find the derivative of: }\:y \:=\: -\csc x - \sin x$

$\text{I got: }\:y' \:=\:\csc x\cot x - \cos x$

$\text{However, the answer in the book is: }\,\cos x \cot^2\!x$

$\text{So presumably my calculus is wrong, or I didn't simplify enough.}$

Your calculus is correct!

$\csc x\cot x - \cos x \;=\;\csc x\cdot\tfrac{\cos x}{\sin x} - \cos x$

. . . . . . . . . . . . . . . $=\;\cos x\left(\tfrac{\csc x}{\sin x} - 1\right)$

. . . . . . . . . . . . . . . $=\;\cos x\left(\frac{\csc x}{\frac{1}{\csc x}} - 1\right)$

. . . . . . . . . . . . . . . $=\;\cos x\underbrace{\left(\csc^2\!x - 1\right)}_{\text{This is }\cot^2x}$

. . . . . . . . . . . . . . . $=\;\cos x\cot^2\!x$