Question

Find the derivative of the function
y = cot^2(sin θ)

Answer 1

`(dy)/(d\theta)=-2\csc^2(\sin\theta)\cot(\sin\theta)(\cos\theta)`

Explanation:

`\displaystyle y=\cot^2(\sin\theta)\\\\y=(\cot(\sin\theta))^2\\\\(dy)/(d\theta)=(\cot(\sin\theta))'\cdot2\cot(\sin\theta)\\\\(dy)/(d\theta)=-\cos\theta\csc^2(\sin\theta)\cdot2\cot(\sin\theta)\\\\(dy)/(d\theta)=-2\csc^2(\sin\theta)\cot(\sin\theta)(\cos\theta)`

The problem takes some rearranging, but mostly simple with use of chain rule and the fact that
`(d)/(dx)\cot(x)=-\csc^2(x)`.