Difficulty solving ∬ ln ⁡ ( x + y 2 ) ⋅ y 2 d x d y with change of variables

Question

Difficulty solving   ∬  ln  ⁡  (  x  +      y    2    )  ⋅      y    2      d  x    d  y with change of variables

Pranav Greer · Accepted Answer

Let me try to guess. You are interested in a function f:

\frac{\partial^{2}}{\partial x \partial y} f (x, y) = y^{2} \ln (x + y^{2}) .

You attempt to do a substitution

u = x + y^{2}

and find function

g (u (x, y), v (y)) = f (x, y)

\frac{\partial}{\partial y} (\frac{\partial}{\partial x} g (u, v)) = \frac{\partial}{\partial y} (\frac{\partial g (u, v)}{\partial u} \frac{\partial u}{\partial x}) = \frac{\partial}{\partial y} (\frac{\partial g (u, v)}{\partial u}) = \frac{\partial^{2} g}{\partial u^{2}} \frac{\partial u}{\partial y} + \frac{\partial^{2} g}{\partial u \partial v} \frac{\partial v}{\partial y} = (2 y \frac{\partial^{2}}{\partial u^{2}} + 2 y \frac{\partial^{2}}{\partial u \partial v}) g = 2 \sqrt{v} (\frac{\partial^{2}}{\partial u^{2}} + \frac{\partial^{2}}{\partial u \partial v}) g .

So we see that operator

\frac{\partial^{2}}{\partial x \partial y}

doesn't transform into

\frac{\partial (u, v)}{\partial (x, y)} \frac{\partial^{2}}{\partial u \partial v}

the same way as dxdy transforms to

\frac{\partial (x, y)}{\partial (u, v)} d u d v

. Whereas in 1D

\frac{\partial}{\partial q}

is indeed

\frac{\partial p}{\partial q} \frac{\partial}{\partial p}

. This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.

Difficulty solving ∬ ln ⁡ ( x + y 2 </ms