Question

Find f' of f(x)=sin(2x)^2 at f'(pi/6)

Answer 1

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

The derivative of f(x) is equal to f'(x)=(d)/(dx) sin(2x)^(2).
f'(x)=(d)/(dx) sin(2x)^(2)

Remove the parentheses around the expression cos(v).
(d)/(dv) sin(v)=cos(v)

Simplify the derivative.
(d)/(dx) sin^(2)(2x)=4sin((2x))cos((2x))

The derivative of the function is 4sin((2x))cos((2x)).
f'(x)=4sin((2x))cos((2x))

Answer 2

f'(pi/6)=4sin((2[pi/6]))cos((2[pi/6]))

Answer 3

COuld you look at this Prove that 76^(130) + 1 is not prime

Answer 4

I got to go now , you can tell others to solve that

Answer 5

ok thanks though