Derivative help

[k_rock923]

I know that the derivative os cos(4x) is -4sin(4x). . .so does that mean that you just take the four out and put it in front of the derivative of cos?

And i can't figure out for anything how to do 3xsin(3x)

[cyu]

(3x)(derivative of sin(3x))+(sin3x)(3)

I took calc last year.

[k_rock923]

that doesn't help because i get stuck on derivative of sin(3x)

edit/ is 3sin(3x) + 9xcos(3x) the final answer?

[cyu]

Yep. looks right.

[griffinn]

d (3x sin 3x) / dx
= [d(3x) / dx] * [d (3x sin 3x) / d(3x)]
= 3 * { 3x [d(sin 3x)]/d(3x) + sin 3x * [3x/d(3x)] }
= 3 * ( 3x cos 3x + sin 3x )
= 9x cos 3x + 3 sin 3xSo yeah, it's correct.

[ElChevelle]

I concur.
:p

[k_rock923]

something like x(2sinxcosx) becomes 2xsinxcosx right? why is only the 2 multiplied by the x?

[griffinn]

That's the associative property of multiplication. Basic algebra.

[deeder7001]

i don't remember how to do advanced math anymore. it's been about a year since i've been in a math class.

[k_rock923]

thanks guys. once again, basic algebra ruins an otherwise correct problem

[cyu]

Originally posted by k_rock923
something like x(2sinxcosx) becomes 2xsinxcosx right? why is only the 2 multiplied by the x?
are you talking about the derivative of that?

[k_rock923]

no. that was part of the answer to xsinsquaredx
so it was x(2sinxcosx)+(sin squared x)1
and that turned into 2xsinxcosx + sin^2 x

if that's wrong, please tell me.

[ryan]

Ask in the AVS forums, they could most likely help.

[griffinn]

Sssshhhh... he's got it right, but we're not supposed to tell him!

/me hides

[Hollow]

k_rock, look up the chain rule in your book. If you don't get it from the book, find someone to show you how it works until you get it. Have random people tell you if you got one question right without understanding the mechanism is wasting your time and ours.

[k_rock923]

hollow, thats how i solved it. i'm just making sure that i did it right. thanks though.