If f(x)=sin^2(4x),f'(pi/12)=?

Question

please help me

Answer 1

f'(x)=2sin4x*4cos4x
when x=pi/12
8 sin pi/3cospi/3
=8*1/2*rt3/2=2rt3

Answer 2

Well, f' will be 4 * 2sin(4x)*cos(4x), and f'(pi/12) would be

4 * 2sin(pi/3)*cos(pi/3) = 4 * 2 * sqrt(3)/2 * 1/2 = 2 * sqrt(3)

Answer 3

f(x) = (sin(4x))^2

using www.quickmath.com

f'(x) = 8cos(4x)sin(4x)
f'(x) = 8 * cos(4(pi/12)) * sin(4(pi/12))
f'(x) = 8 * cos(pi/3) * sin(pi/3)
f'(x) = 8 * (1/2) * (sqrt(3)/2)
f'(x) = 4 * (sqrt(3)/2)
f'(x) = 2sqrt(3)

if you want to go into detail, then

cos(4x) = sin(x)^4 - 6cos(x)^2sin(x)^2 + cos(x)^4
sin(4x) = -4cos(x)sin(x)^3 + 4cos(x)^3sin(x)

Answer 4

hi there! haha yes trigo sucks. but here's the solution..

f(x)=sin^2(4x)=(sin4x)^2

therefore,

f'(x)=8cos4xsin4x

to find f'(pi/12), sub x=pi/12

therefore,
f'(pi/12)=8cos(pi/3)sin(pi/3)
=8(0.5)(root3/2)
=2(root3)

((:

Answer 5

I study it in my school too,stop it please!:)