Question

What is the first derivative of `sin^2(6x)`?

Answer 1

`(dy)/(dx)=12sin6xcos6x`

Explanation:

we will need the chain rule

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

`y=sin^2(6x)`

`u=sin6x=>(du)/(dx)=6cos6x`

`y=u^2=>(dy)/(du)=2u`

`:.=>(dy)/(dx)=2uxx6cos6x`

`(dy)/(dx)=12ucos6x=12sin6xcos6x`