Question

Question #b5fef

Answer 1

`f'(x)=sqrt3`

Explanation:

.

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

Let `u=2x`

`du=2dx`

`f(x)=sin^2u`

Let `z=sinu`

`dz=cosudu`

`f(x)=z^2`

`f'(x)=2z`

Now, we substitute back for `z`:

`f'(x)=2sinucosu`

Now, we substitute back for `u`:

`f'(x)=2sin(2x)cos(2x)(2)`

`f'(x)=4sin2xcos2x`

`f'(pi/6)=4sin((2pi)/6)cos((2pi)/6)=4sin(pi/3)cos(pi/3)=4sqrt3/2(1/2)`

`f'(x)=sqrt3`

Answer 2

See below.

Explanation:

Find derivative of `f(x)=sin^2(2x)`

Using chain rule:

`d/dx(sin^2(2x))=2(sin(2x))*2cos(2x)=4sin(2x)cos(2x)`

`:.`

`f'(pi/6)=4sin(pi/3)cos(pi/3)=4(sqrt(3)/2)(1/2)=sqrt(3)`