If all the partial derivatives of $f$ in a neighbourhood of $a$ exist and are continuous at $a$ then $f$ is differentiable at $a$

Suppose $$f(x,y)= \begin{cases} \frac{y^3}{x^2+y^2}, & (x,y)=(0,0)\\ 0, & (x,y)=(0,0) \end{cases}$$

(First, we observe that $f$ is continuous at $(0,0)$)

The partial derivatives are:

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

$$\dfrac{\partial f}{\partial y}=\frac{3y^2(x^2+y^2)-2y^4}{(x^2+y^2)^2}$$

which exist for sure in $\Bbb R^2 \setminus \{(0,0)\}$ and are continuous there.

Now if they also exist at $(0,0)$ and are continuous at $(0,0)$ then $f$ is differrentiable at $(0,0)$ as stated by the squared proposition.

My question is, what exactly is happening with these partial derivatives at $(0,0)$? Do they exist? What is their value there? Do I just set their value to be $0$ or..??


1 Answer 1



check from definition

$$\frac{\partial f(0,0)}{\partial x}=\lim_{\Delta x \to 0} \frac{f(\Delta x, 0)-f(0,0)}{\Delta x}=\lim_{\Delta x \to 0} \frac{f(\Delta x, 0)}{\Delta x}$$

to verify if it exists.

  • $\begingroup$ So they do exist, and they are zero. Now to check the differentiability of $f$ we need to check their continuouity at zero. Many thanks! $\endgroup$ Nov 8, 2017 at 8:57
  • $\begingroup$ Is there a different way to check if this partial derivative exists at zero and calculate it? $\endgroup$ Nov 8, 2017 at 14:46
  • $\begingroup$ I am not sure of the alternative method. I thought checking from definition is easy for this case? $\endgroup$ Nov 8, 2017 at 18:16

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