Question

How do you find the derivative of `sin ^2 (2x) + sin (2x+1)`?

Answer 1

You can find it like this:

Explanation:

`f(x)=sin^2(2x)+sin(2x+1)`

Applying the chain rule successively:

`f'(x)=2[sin(2x)]^1xxcos(2x)xx2+[cos(2x+1)xx2]`

`:.f'(x)=4sin(2x)cos(2x)+2cos(2x+1)`