Question

What is the derivative of `tan(2x)`?

Answer 1

`(dy)/(dx)=2sec^2(2x)`

Explanation:

`y=tan(2x)`

`u=2x=>(du)/dx=2`

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

the chain rule:

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

`:.(dy)/(dx)=(sec^2u)xx2`

`(dy)/(dx)=2sec^2u=2sec^2(2x)`