Derivatives of Trigonometric Function

[PMC_l0ver]

Homework Statement

y=(sin2x)(cos2)

Homework Equations

Product Rule for Derivatives
identities:
derivatives of
Sinx = Cosx
Cox = -Sinx

The Attempt at a Solution

i used the product and chain rule for derivatives then do the identities

y = sin2x*cos2x
dy/dx = (Cos2x)(2) (Cosx) + (-Sin2x)(2)(Sin2x)
dy/dx = 2(Cos2x^2) - 2(Sin4x^2)

i know that my answer is wrong... I am kinda confuse about multiplying them
can please anyone help me

 

[micromass]

I assume that your problem is y=(sin 2x)(cos 2x), you made a little type.

Then the derivative is 2(cos2x)(cos2x)+2(-sin 2x)(sin 2x)=2cos²2x-2sin²2x

 

[PMC_l0ver]

I assume that your problem is y=(sin 2x)(cos 2x), you made a little type.

Then the derivative is 2(cos2x)(cos2x)+2(-sin 2x)(sin 2x)=2cos²2x-2sin²2x

yes its y= (sin 2x) (cos 2x)
however
ur answer does nnot match to the answer key
and also this part u did " 2(-sin 2x)(sin 2x) " isn't suppose to be -2(Sin²4x)

btw the answer base on the book is

2cos4x and I am trying to figure out how to do it properly
btw thanks for trying my problem i appreciate it!

 

[micromass]

Ah yes. You have the formula

\cos(2x)=\cos^2(x)-\sin^2(x)

So my formula

2\cos^22x-2\sin^22x = 2\cos (2(2x))=2\cos(4x)

Note that it isn't true that 2cos(2x)=cos(4x). I say this, because it seems that you did that...

 

[PMC_l0ver]

can you please explain it steps by steps
im sorry because I am still new at this lesson and i can't do it in my head yet
thank you!
the part that i get confuse would probably be multiplying trigo

 

[micromass]

So

[\cos(2x)\sin(2x)]^\prime=2(\cos 2x)(\cos 2x)+2(-\sin 2x)(\sin 2x)=2\cos^22x-2\sin^22x

Using the trigoniometric formula \cos(2\alpha)=\cos^2\alpha-\sin^2\alpha with \alpha=2x, we get

2\cos^22x-2\sin^22x=2(\cos^22x-\sin^22x)=2\cos(2(2x))=2\cos(4x)

Also note that it is NOT true that sin(x)sin(y)=sin(xy). I read this in your first post. It is NOT TRUE, it is EVIL!

 

[PMC_l0ver]

because the answer says

Since y=1/2 sin (4x), it follows that dy/dx = 1/2 cos 4x (4) = 2cos4x

i need to know how to get in this part
your calculation looks good but I am not sure where it came from

 

[micromass]

The answer uses the formule \sin(2\alpha)=2\sin(\alpha)\cos(\alpha).

So instead of deriving cos(2x)sin(2x), they derived (1/2)sin(4x).

 

[PMC_l0ver]

ohhh ok i see thank you!
i totally forgot about this trigonometric formula
thank you so much! it refreshed everything!