Question

What is the derivative of `sin^2(4x)`?

Answer 1

`(dy)/(dx)=8sin(4x)cos(4x)`

Explanation:

`y=(sin(4x))^2=(f(x))^2`

`y=(f(x))^n=>(dy)/(dx)=n*f'(x)*(f(x))^(n-1)`

So, `y=(sin(4x))^2`
`(dy)/(dx)=2*sin(4x)*d/(dx)sin(4x)`
`=2sin(4x)*4cos(4x)`
`=8sin(4x)cos(4x)`