Question

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!

Answer 1

`(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)`.