partial derivation 2nd order for a function defined by parts

Question

Do this contradict Clairaut? How to do c) and whats the others results just to confirm if i am right. Thanks

Answer 1

If $(x,y) \neq (0,0)$, to compute $f_x(x,y)$ we treat $y$ as constant and take the derivative w.r.t. $x$, using the quotient rule. We have $$f_x(x,y) = \frac{(3x^2y - y^3)(x^2 + y^2) - (x^3y - xy^3)(2x)}{(x^2 + y^2)^2}$$ To calculate $f_y(x,y)$ its the same thing. You can use Wolfram Alpha, for instance, to check. From here on, you have to use the definition of the partial derivatives. Again, I'll do one of them. $$\begin{align} f_x(0,0) &= \lim_{h \to 0} \frac{f(0 + h, 0) - f(0,0)}{h} \\ &= \lim_{h \to 0} \frac{\frac {h^3 0 - h0^3}{h^2 + 0^2} - 0}{h} \\ &= \lim_{h \to 0} \frac{0}{h} = 0 \end{align}$$ It is very important that you understand the definition. This way you can solve problems like this and analyze complicated points like $(0,0)$ in your function $f$. To calculate second order mixed derivatives on that point, we use the definition again. Here I prefer to use Leibniz's notation, to make clear which variable we are taking derivatives w.r.t.. I'll compute $f_{yx}(0,0) = \frac{\partial^2 f}{\partial y \partial x}(0,0)$. Now: $$\begin{align} \frac{\partial}{\partial y} \left(\frac{\partial f}{\partial x} \right) (0,0) &= \lim_{k \to 0} \frac{\frac{\partial f}{\partial x}(0,0+k) - \frac{\partial f}{\partial x}(0,0)}{k} \\ &= \lim_{k \to 0} \frac {\frac {(3.0^2 k - k^3)(0^2 + k^2) -(0^3 k - 0 k^3)(2 \cdot 0)}{(0^2 + k^2)^2} - 0}{k} \\ &= \lim_{k \to 0} \frac{\frac{-k^5}{k^4}}{k} \\ &= \lim_{k \to 0} -1 = -1 \end{align}$$

You do the other one to practice. Always play close attention to what gets plugged where. In the first line of every calculation I did, I did it like this to help you see what is happening.

Important: I got $-1$ instead of $1$ like your text says, because of a matter of convention. Your text must mean $f_{yx}(x,y) = \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) (x,y)$, opposing what I had interpreted. Again, I am not wrong, and neither is your text. I had not counted on this problem, but now it doesn't matter, more information to you. I hope you really don't get confused by this. And the mixed partials being different does not contradict Clairaut-Schwartz, because the derivatives aren't continuous. You can see for yourself, based on my calculations above that probably $\lim_{(x,y) \to (0,0)} f_x(x,y) \neq f_x(0,0)$. If this limit actually works, then the analogous with $f_y(x,y)$, or $f_{xy}, f_{yx}$ certainly fails. I hope this makes things clearer to you. Good studies (: