Quick math problem. Reward included.
 

I just recently enrolled in a math class for college and it's been a year since I took any type of math since I basically finished calculus in highschool and forgot almost everything in that class. Well, apparantely this class jumps right into multivariable calculus and expects you to know single variable calculus like the back of your hand. Needless to say I'm pretty screwed.

Anyways, the first person to solve this problem correctly while giving me a clean step by step manner will get $20 shipped to pokerstars immediately. Is this cheating? Eh, sure, but theres tons of problems like this and I'm just asking for a basic overview of how to do one. Anyways the problem is:

Compute dy/dx if y = cos(3x)e^(-x^2).

I know vaguely I have to use some kind of chain rule here as well as some funny business with e where it's basically repeates itself but honestly I'm completely lost.

Re: Quick math problem. Reward included. *DELETED*
 

Post deleted by rivermetimbers

Re: Quick math problem. Reward included.
 

e^(-x^2)*((4x^2 - 11)cos(3x)+12xsin(3x))
... reward?????

Re: Quick math problem. Reward included.
 

You'll need to perform the product rule.

It will be the derivative of cos(3x) multiplied by e^(-x^2) plus the derivative of e^(-x^2) multiplied by cos(3x).

So we get -3sin(3x)e^(-x^2) - 2xcos(3x).

Edit: SN = TurtlePiss

Re: Quick math problem. Reward included.
 

for step by step, message me on AIM at shmu529

Re: Quick math problem. Reward included.
 

damn it, some guy at nwp answered it pretty clearly. but as a sign of good gesture, i'll send you 10 too turtle.

Re: Quick math problem. Reward included.
 

not me?

Re: Quick math problem. Reward included.
 

[ QUOTE ]
I just recently enrolled in a math class for college and it's been a year since I took any type of math since I basically finished calculus in highschool and forgot almost everything in that class. Well, apparantely this class jumps right into multivariable calculus and expects you to know single variable calculus like the back of your hand. Needless to say I'm pretty screwed.

Anyways, the first person to solve this problem correctly while giving me a clean step by step manner will get $20 shipped to pokerstars immediately. Is this cheating? Eh, sure, but theres tons of problems like this and I'm just asking for a basic overview of how to do one. Anyways the problem is:

Compute dy/dx if y = cos(3x)e^(-x^2).

I know vaguely I have to use some kind of chain rule here as well as some funny business with e where it's basically repeates itself but honestly I'm completely lost.

[/ QUOTE ]

Re: Quick math problem. Reward included.
 

dy/dx if y = cos(3x)e^(-x^2).

okay so this is a chain rule problem.

so break this into two functions: g(x)=cos(3x) and f(x)=e(-x^2).

derivative of cosine function (cos(ux)) = -u sin(ux)
derivative of exponential function (e(ux))=u*e(ux)

so derivative of g(x) and f(x) are:

dg/dx=-3*sin(3x) df/dx=-2x*e(-x^2)

Chain rule basically says you can take the derivate of one function while holding the other constant and then take the derivative of the other function while holding the other one constant.

derivative: dy/dx= g(x)*df/dx+f(x)*dg/dx

g(x)=cos(3x) and f(x)=e(-x^2).
dg/dx=-3*sin(3x) df/dx=-2x*e(-x^2)

substituting in:
dy/dx=cos(3x)*-2xe(-x^2)+e(-x^2)*-3*sin(3x)

simplified: -e(-x^2)*[2x*cos(3x)+3sin(3x)]