Question #b5fef

2 Answers
Dec 18, 2017

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

Dec 18, 2017

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)