Question

Hey does anyone know how to get wolfram to compute all the second derivatives?

I have my function:

f(x,y)=(x^2ye^(-x^2-y^2))

I want to compute the second partial derivatives wrt x, y and the mixed. Just to check my work. I can only seem to get it to show me the second partial wrt x

Answer 1

you could try this: http://www.wolframalpha.com/input/?i=d%5E2%2Fdy%5E2+%28x%5E2ye%5E%28-x%5E2-y%5E2%29%29

Answer 2

thanks a lot I tried that so many times but it's the brackets that did it. still cant figure out the mixed though

Answer 3

what exactly do you mean by "the mixed"?

Answer 4

do you mean this: http://www.wolframalpha.com/input/?i=%28d%5E2%2Fdy%5E2+%28x%5E2ye%5E%28-x%5E2-y%5E2%29%29%29+%2B+%28d%5E2%2Fdx%5E2+%28x%5E2ye%5E%28-x%5E2-y%5E2%29%29%29

Answer 5

Uh I don't think I mean that I mean \[d^2/dxdy \]

Answer 6

is that the same as d/dx + d/dy?

Answer 7

or is it d/dx followed by d/dy?

Answer 8

the second case can be done like this: http://www.wolframalpha.com/input/?i=d%2Fdy+%28d%2Fdx+%28x%5E2ye%5E%28-x%5E2-y%5E2%29%29%29

Answer 9

awesome thanks. And yeah it is d/dy followed by d/dx which in this case is the same as what you showed since the function is exact.

just so you know, every function of two variables will have four second partial derivatives \[f_{xx}, f_{yy} , f_{xy}, and, f_{yx} \] and the later two are commonly referred to as "mixed" and as I said before usually are the same.

Answer 10

I've learnt something new today! - thanks for that explanation :)