Question

derivative of y =cot^2(sinx)

Answer 1

use chain rule

u = sin x
du = cos x

now find
\[\frac{d}{du} \cot^2 u\]
\[=2 \cot u (-\csc^2 u)\]
plug u back in and multiply by du
\[\rightarrow -2 \csc^2 (\sin x) \cot (\sin x) \cos x\]