How to differentiate y=xe to the power of x in d square y over dx square?

Question

jack1 · Answer

$$y=xe ^{x}$$

jack1 · Answer

in what exactly?

imtant · Answer

yes the first one.

jack1 · Answer

$$y=xe^x \frac{ d^2y }{ dx^2 }$$

jack1 · Answer

so is this it?

imtant · Answer

ans. in second differentiate.

imtant · Answer

what I mean is this, in d square y over dx square.

jack1 · Answer

so find 2nd derivative of y = x e^x ... yeah?

imtant · Answer

Yes.

jack1 · Answer

ok
y' = (x+1) e^x

jack1 · Answer

y'' = (x+2) e^x = 2nd derivative of y with relation to x

imtant · Answer

How do you get the first derivative?

jack1 · Answer

ok the derivative of e^x = e^x
and the derivative of x is 1
so using the product rule:
d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = e^x and v = x

you end up with y' = (x+1) e^x

jack1 · Answer

then do the same again on the derivative of y to get y''

jack1 · Answer

i gotta go dude, sorry @Directrix  could you take over please man?

imtant · Answer

It's okay Jack1.Thank you vm.

directrix · Answer

@Imtant   I don't know where you and Jack1 left off.

For the second derivative, did you get  2*e^x + x*e^x ?

imtant · Answer

No,I don't know how to do it.

directrix · Answer

Did you agree with this statement written by Jack1? -->    y' = (x+1) e^x

imtant · Answer

I got for first derivative is e^x+xe^x.