PARTIAL DERIVATIVE: Compute (del^2g)/(delxdely) and (del^2g)/(delydelx) for the function g(x,y)=(1+y^2)e^(x^2-y) and show that they are both equal?

Solution to this problem is 2xe^(x^2-y)(2y-1-y^2). Please, could someone explain this thoroughly to me and future viewers? Thanks!

1 Answer
May 29, 2018

(del^2g)/(delxdely)=(del^2g)/(delydelx)=2xe^(x^2-y)(2y-1-y^2)

Explanation:

For g(x,y)=(1+y^2)e^(x^2-y), we want to g_(xy) and g_(yx) and, in particular, show that they are equal.

In order to find g_(xy), we need to first find g_x and the find the partial derivative of g_x wrt to y. To find the partial derivative wrt x of g, we treat y as a constant and simply differentiate g with respect to x.

(delg)/(delx)=(1+y^2)/e^yd/dxe^(x^2)=2xe^(x^2-y)(1+y^2)

Now we do take the derivative of 2xe^(x^2-y)(1+y^2) wrt y.

(del^2g)/(delydelx)=2xe^(x^2)d/dye^(-y)(1+y^2)=2xe^(x^2)(2ye^-y-e^-y(1+y^2))=2xe^(x^2-y)(2y-1-y^2)

Now, we could do the same thing in the opposite order (perhaps you want to as an exercise), but there is a theorem stating that if g_x and g_y are continuous for a function g(x,y), then we know that g_(xy)=g_(yx). It is clear that all of these functions are continuous, and g_y is also continuous, so we know that g_(xy)=g_(yx).