# Prove that the function is differentiable at (0,0)

The function is shown below. Its not a very complicated function. $$f(x,y)=\sqrt{9-x^2-y^2}$$ I was wondering is it sufficient to say that since $f_x(0,0)$ and $f_y(0,0)$ are both zero and since $f_x$ and $f_y$ are continuous (by finding the limit as it appraoches to $(0,0)$ for both..is there a faster way?) at $(0,0)$ therefore $f(x,y)$ is differentiable at $(0,0)$. Are there any other ways.

While browising around other questions, this limit came up twice: $\lim_{(x,y)\to (0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0$. Is it applicable to all functions? I doubt it because the denominator is specific to the question. I would appreciate alternative methods and any help.

• You could prove directly that $$\lim_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=0,$$ but I think that your approach is correct. – Siminore Nov 2 '13 at 11:35

Now, to the problem. Youhave to show that the differential of $f$ at $(0,0)$ is the zero-linear function, that, is, $$\lim_{(x,y)\rightarrow (0,0)}\dfrac{f((0,0)+(x,y))-f(0,0)-\mathbf{0}(x,y)}{\Vert (x,y)\Vert}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{\sqrt{9-x^2-y^2}-3}{\Vert (x,y)\Vert}=0$$
You can calculate this limit directly, and attain the result you want (you can use the fact that $\lim_{h\rightarrow 0}\dfrac{\sqrt{9-h^2}-3}{h}=0$, in one-variable calculus!).
Once you have shown that the function is continuous, you can certainly prove that the function is differentiable by showing that the partial derivatives agree. Indeed you have that the derivative of $f$ in direction $\mathbf{v}$ (a unit vector) at $(x,y)$ is given by $$(\nabla f)_{(x,y)}\cdot\mathbf{v}$$ (you can try to show this: simply apply a rotation to the coordinate frame).
Another way is to compute directly the derivative, which gives you $$\lim_{\|\mathbf{v}\|\rightarrow 0}\frac{f(x+v_x,y+v_y)-f(x,y)}{\|v\|}$$ Notice that in the specific case $(x,y)=(0,0)$ this gives you exactly the formula written by @Siminore.