How do you find derivative of xe^(x^2)

[Lo.Lee.Ta.]

1. This may seem like a dumb question, but I don't really know the steps of how to find this derivative...

f(x)= xe^x2

2. The answer is supposed to be: e^-x2*(1-2x²)

I thought to do u-substitution.

u= x²

du= 2xdx

1/2du = xdx

(1/2)e^udu

(1/2)ue^u-1

(1/2)(x²)(e^x2-1)

hhhh =_= How is the real answer supposed to be: e^-x2(1-2x²)?

Thanks :)

 

[jedishrfu]

you need to apply the product rule y=uv dy/dx = du/dx * v + u * dv/dx

and for the dv/dx you need to use the dy/dx = e^u du/dx

There's also some confusion here: y = x * e ^ (x^2) or is it y = x * e ^ (-x^2)

 

[Lo.Lee.Ta.]

The product rule for this would be:

(1)(e^x2) + (x)(2xe^x2)

= e^x2 + 2x²e^x2

If I factor out e^x2, I would get:

e^x2(1 + 2x²)

Similar, but they have negatives! How is that? :?

 

[Infrared]

Is it possible that the problem was supposed to have $e^{-x^2}$? That is the only explanation I can see.

 

[Lo.Lee.Ta.]

Hey, you were right! I copied it down wrong! #O_O

It was xe^-x2

e^-x2(1-2x²)