Calculus I Derivatives of Trigonometric Functions help

[Reggie5489]

To keep from going on and on I'm going to just post pics of the problem and work I have done. I have had this explained to me a few times but in math new topics take a little longer for me to digest.

http://tinypic.com/view.php?pic=4ljaea&s=6

The original problem is 1/4 sin^2 2x.

I first rewrote the problem as 1/4 multiplied by sin(2x)^2. then I attempted to differentiate the sin(2x) and go from there...the answer should be sin(2x) cos(2x) equaling 1/2 sin(4x). However as with the rest of the trig problems I am becoming stumped. I believe there is a chain rule inside of a chain rule somewhere but I am not sure. Any help is gladly appreciated, I really have been trying to figure these equations out for the majority of the day. I am spending hours on these trig problems that should only be taking me minutes to solve.3

 

[daon2]

That is a trig identity. sin(2y) = 2sin(y)cos(y). So sin(4x) = sin(2*2x) = 2sin(2x)cos(2x)

 

[soroban]

Hello, Reggie5489!

\(\displaystyle \text{Differentiate: }\:f(x) \:=\:\frac{1}{4}\sin^2(2x)\)

\(\displaystyle \text{Answer: }\:f'(x) \;=\;\sin(2x)\cos(2x) \;=\;\frac{1}{2}\sin(4x)\)

\(\displaystyle \text{We have: }\:f(x) \:=\:\frac{1}{4}\big[\sin(2x)\big]^2\)

\(\displaystyle \text{Then: }\:f'(x) \:=\:\frac{1}{4}\cdot 2\big[\sin(2x)\big]\cdot\cos(2x)\cdot 2\)

. . . . . . . . . . \(\displaystyle =\;\sin(2x)\cos(2x)\)

. . . . . . . . . . \(\displaystyle =\;\frac{1}{2}\big[2\sin(2x)\cos(2x)\big]\)

. . . . . . . . . . \(\displaystyle =\;\frac{1}{2}\sin(4x)\)