Question

How do you find the derivative of `1/2sin2x`?

Answer 1

`(dy)/(dx)=cos2x`

Explanation:

we will use the chain rule

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

`y=1/2sin2x`

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

`color(red)(y=1/2sinu=>(dy)/(du)=1/2cosu)`

`(dy)/(dx)=color(red)(1/2cosu ) xxx2`

`(dy)/(dx)=cosu=cos2x`