Differentiating f(x) = x·e^(-π·x²)

[galipop]

f(x) = x.e^(-pi.x^2)

this is how I tried to do it...

f'(x) = x.(-pi.2.x.e^(-pi.x^2)) + 1.e^(-pi.x^2)
f'(x) = (1 - 2.pi.x^2).e^(-pi.x^2)

can anyone see anything wrong with this?

Cheers

 

[arildno]

Seems correct to me, at least

 

[NateTG]

Well, you could try this way:
$$x \times e^{ - \pi x^2}=e^{- \pi x^2 \ + ln x}$$
so you get:
$$(\frac{1}{x} - 2\pi x) e^{- \pi x^2 \ + ln x}$$
which is the same.

Looks good to me.