Question

True or False:

The mixed partial derivatives $\displaystyle{\frac{\partial^2 z}{\partial y \partial x}}$  and $\displaystyle{\frac{\partial^2 z}{\partial x \partial y}}$   are equal

Answer 1

Answer is false.

The mixed partial derivatives $\frac{\partial²Z}{\partial{x}\partial{y}} and \frac{\partial²Z}{\partial{y}\partial{x}}$ are not necessarily equal. If they are continuous at a point then equal at that point. For example let us consider a piece wise defined function,

$Z = \begin{cases} xy\frac{x²-y²}{x²+y²} &\text{if } (x,y)≠(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$

=> $Z = \begin{cases} \frac{x³y-xy³}{x²+y²} &\text{if } (x,y)≠(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$

Here the symbols $f_{x} , f_{y}, f_{xy}and f_{yx}$ are used for partial and mixed partial derivatives

$f_{x}=\begin{cases} \frac{(x²+y²)(y³-3x²y)-(xy³-x³y)2x}{(x²+y²)²}&\text{if } (x,y)≠(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$

$f_{x}=\begin{cases} \frac{(3x²-y³)}{x²+y²}-\frac{2x(x³y-xy³)}{(x²+y²)²})&\text{if } (x,y)≠(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$ Similarly

$f_{y}=\begin{cases} \frac{(x³-3xy²)}{x²+y²}-\frac{2y(x³y-xy³)}{(x²+y²)²})&\text{if } (x,y)≠(0,0) \\ 0 &\text{if } (x,y)=(0,0) \end{cases}$ So

$f_{xy}(0,0)= lim_{h->0}\frac{f_{x}(0,h)-f_{x}(0,0)}{h}= lim_{h->0}\frac{-h³}{h²h}=-1$

$f_{yx}(0,0)= lim_{h->0}\frac{f_{y}(h,0)-f_{y}(0,0)}{h}= lim_{h->0}\frac{h³}{h²h}=1$

$(\frac{\partial²Z}{\partial{x}\partial{y}})_{(0,0)} = -1 and (\frac{\partial²Z}{\partial{y}\partial{x}} )_{(0,0)} = 1$

So $\frac{\partial²Z}{\partial{x}\partial{y}} ≠\frac{\partial²Z}{\partial{y}\partial{x}}$