Question

How do you find the derivative of `y=sin^ntheta`?

Answer 1

`= n*sin^n theta* cot theta`

Explanation:

`y = sin ^n theta`

let say `x = sin theta`,
`(dx)/(d theta) = cos theta`

`y = x^n`
`(dy)/(dx) = n*x^(n-1) = n sin^(n-1) theta`

therefore,
`(dy)/( d theta) =(dy)/(dx) * (dx)/(d theta)`

`= n*sin^(n-1) theta* cos theta = n*sin^n theta / sin theta* cos theta`

`= n*sin^n theta * cot theta`