Question

Derivative of x^3cos3x

Answer 1

You got a formula for the product rule, it shouldn't be too hard.\[fg=f'g+fg'\] for reference though the answer should be:\[-3x^3\sin(3x)\]

Answer 2

sorry, that should be derivative of fg in that formula i put. :P

Answer 3

Thanks I understand the product rule just idk how to find the derivative of cos3x. Would it be -sin3x? @Mikogekz

Answer 4

Close, but not quite:\[d/dx(\cos(f(x))=-f'(x)\sin(f(x))\]by the chain rule, so your answer would be:\[-3\sin3x\]

Answer 5

It works the same for sin(x) and tan(x) also. If you got the trig function of something and you take that derivative, always put the derivative of what is in the function out the front of the function and multiply.

Answer 6

Ok so product rule would be 3x^2*cos3x-3sin3x I'm sorry it's just I'm not sure how u got ur answer @Mikogekz

Answer 7

Oh... I only wrote the second part of it, oh well, you're almost right with your answer. Just make sure where you have 3sin3x you don't forget to multiply by the other function x^3. It seems you've just taken the derivative and forgotten the other part. For reference THE REAL CORRECT ANSWER SHOULD BE:\[3x^2\cos(3x)-3x^3\sin3x\]p.s. sorry about that mistake earlier. :)

Answer 8

It's ok thanks so much(:@Mikogekz

Answer 9

But that's weird it's not an answer choice I have two similar : 3x^2(cos3x+xsin3x) or 3x^2(cos3x-xsin3x)

Answer 10

@Mikogekz

Answer 11

It would be the second of those 2 answer. You can just factorize the expression and take out a common factor of 3x^2. Or, you can just expand the answers to show that it is the same as the one you have.