Proving that the function $f(x,y)=\frac{x^2y}{x^2+y^2}$ has partial derivatives at $0$.

Question

Let:

$$f: (x,y) \mapsto \frac{x^2y}{x^2+y^2}$$ and $\:f(0,0)=0$

It's partial derivatives are:

$$\frac{\partial f}{\partial x} (x,y)=\frac{2xy^3}{(x^2+y^2)^2}$$

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

How can I prove that is has partial derivatives at $0$ and calculate them? Would showing that $f$ is continues at $0$ and stating that the partial derivatives at $0$ are also null suffice? Or should I try showing that the partial derivatives are equal at $0$?

Answer 1

We have that $\frac{\partial f}{\partial x}(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}{x}=\lim_{x\to0}\frac{0-0}{x}=0$, and similarly for $\frac{\partial f}{\partial y}(0,0)$.

Answer 2

Going by the definition the partial derivative in $x$ at $(0,0)$ is $\displaystyle \lim_{h \to 0}\frac{f(h,0) - f(0,0)}{h} = \displaystyle \lim_{h \to 0}\frac{0 -0}{h} =0.$

Can you try the partial derivative in $y$ now?

Answer 3

Hint: partial derivatives of $f$ at zero are ordinary derivatives of $g(x)=f(x,0)$ resp. $h(y)=f(0,y)$ at zero, i.e. $g'(0)$ and $h'(0)$. Can you say how $g$ and $h$ look like near zero?