Consider a function $$f(x,y)= \begin{cases} xy, & |x|<|y| \\ 0, & \mathrm{else} \end{cases} $$ Let's say, we are going to compute $f_{,x}(0,0), f_{,y}(0,0), f_{,xy}(0,0)$ and $f_{,yx}(0,0)$ in point $(0,0)$. Important note: I took the liberty of physical/relativistic notation of partial derivatives $\frac{\partial f}{\partial x}$ as $f_{,x}$. How do I do that? Actually, this is an example of non-interchangeability of second derivatives, my maths teacher has this in his book, stating that $$f_{,xy}(0,0)=0\, \mathrm{and}\, f_{,yx}(0,0)=1$$ however, I do not quite understand how he got into a conclusion that it should be this way(there is no commentary on computing it). I presumed, that the condition stays even after the process of differentiation and since $x=y=0$, then those results should be the same. Am I wrong? Anyway, we can consider also a function $$f(x,y)=\begin{cases}xy\sin(x^2-y^2)^{-1}, & \mathrm{for}\, (x,y)\neq(0,0)\\ 0, & \mathrm{for}\, (x,y)=0 \end{cases}$$ and compute $f_{,x}(0,0), f_{,y}(0,0), f_{,xy}(0,0)$ and $f_{,yx}(0,0)$ where they exist. How do I do that?
1 Answers
You need to go back to definitions. You need \begin{align*} \frac{\partial f}{\partial x}(0,0) &= \lim_{h\to 0}\frac{f(h,0)-f(0,0)}h = 0 \\ \frac{\partial f}{\partial y}(0,0) &= \lim_{k\to 0}\frac{f(0,k)-f(0,0)}k = 0 \\ \frac{\partial f}{\partial x}(0,y) &= \frac{\partial (xy)}{\partial x} = x \quad (y\ne 0) \\ \frac{\partial f}{\partial y}(x,0) &= \frac{\partial (0)}{\partial x} = 0. \end{align*} Now write down $$\frac{\partial^2 f}{\partial x\partial y}(0,0) = \frac{\partial }{\partial x}\Big|_{x=0}\left(\frac{\partial f}{\partial y}(x,0)\right)$$ and $$\frac{\partial^2 f}{\partial y\partial x}(0,0) = \frac{\partial }{\partial y}\Big|_{y=0}\left(\frac{\partial f}{\partial x}(0,y)\right)$$ and compute.
I'll let you attack the second problem now. :)
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\begin{cases} ... \end{cases}dedicated environment, just so that you know. – davidlowryduda Jun 16 '14 at 20:47