Question

How do you differentiate `2cos^2(x)`?

Answer 1

`(dy)/(dx) =-4cosxsinx`

Explanation:

To differentiate `y=2cos^2x` we need the chain rule

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

let `u=cosx=>y=2u^2`

`(du)/(dx)=-sinx`

`(dy)/(du)=4u`

`:.(dy)/(dx)=4uxx(-sinx)`

substitute back for `u`

`(dy)/(dx) =-4cosxsinx`