Question

How do we find the derivative of `sinxcos^2 2x`?

Answer 1

`(dy)/(dx)=cos^3x-2sinxsin4x`

Explanation:

We use here product rule. Here `y=f(x)g(x)` and according to product rule `(dy)/(dx)=(df)/(dx)g(x)+f(x)(dg)/(dx)`

Here we have `f(x)=sinx` and `g(x)=cos^2 2x`

and therefore `(df)/(dx)=cosx` and `(dg)/(dx)=2cos2x xx (-sin2x)xx2=-4sin2xcos2x=-2sin4x`

Hence `(dy)/(dx)=cosx xx cos^2x+sinx xx(-2sin4x)`

= `cos^3x-2sinxsin4x`