Question

How do you find the derivative of `y = sin^2 x`?

Answer 1

`dy/dx=2sinxcosx`

Explanation:

Using `u=sinx` gives us `y=u^2`

`dy/dx=(dy)/(du)*(du)/(dx)`

`(dy)/(du)=2u`
`(du)/(dx)=cosx`

`dy/dx=2ucosx=2sinxcosx`