# How do I prove continuity of partial derivative of $f(x, y) = \sqrt{x^4+y^4}$ at $(0,0)$?

Consider I have $$f(x, y) = \sqrt{x^4+y^4}$$ And I want to check if the function has partial derivatives continuous in point $$(x_0, y_0) = (0, 0)$$

I know theorem, that existence of continuous partial derivatives implies differentiability of this function, and differentiablity implies that function is continuous. I can also test this function for if it is differentiable. Can I use those information to check if partial derivatives are continuous? If not, what is standard method of calculating continuity of partial derivative?

$$f_x=\lim_{x\to0}{\frac{f(0+x, 0) - f(0, 0)}{x}} = 0$$

• You simply showed that $f_x(0,0) = 0.$ You need to show $\lim_{(x,y) \to (0,0)} f_x(x,y) = 0$. – IAmNoOne Jun 4 '14 at 8:37
• Is there a reason I couldn't quickly check partial derivaties continuity to prove differentiability and continuity in the same time? Because there is another method of checking just differentiability. – stil Jun 4 '14 at 8:42
• I thought you were trying to prove that $f_x(x,y)$ is continuous? – IAmNoOne Jun 4 '14 at 8:47
• Yes, but I already can prove differentiability and existence of partial derivatives. I just wanted to know if there is a connection so I could use previous facts to prove $fx$ continuity. – stil Jun 4 '14 at 8:49

You have: $$f_x(x,y)= \begin{cases} \frac{2x^3}{\sqrt{x^4+y^4}}&(x,y)\neq(0,0)\\ 0&(x,y)=(0,0)\\ \end{cases}$$ and this function is continuos in $\mathbb R^2$ because $|f_x(x,y)|\leq 2|x|$ for each $(x,y)\in\mathbb R^2$. By simmetry, also $f_y$ is continuous in $\mathbb R^2$.