Differentiability of the function:$g(x,y)=f(\sqrt{x^2+y^2})$

Question

Let $$g(x,y)=f(\sqrt{x^2+y^2})$$ with $f$ continuously differentiable($\mathbb{R}\to\mathbb{R}$) and $f'(0)=0$.

Then, is $g$ differentiable on $\mathbb{R^2}$? Specifically, does $g(0,0)$ and $f(0)$ have a role to play?

Answer 1

For $g(x,y)$ to be differentiable at $(0,0)$ we have to see if $$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)-\big( g_x(0,0)h+ g_y(0,0)k\big)}{\sqrt{h^2+k^2}}=0$$

We have,

$$g_x(0,0)=\lim _{h\rightarrow 0} \frac{g(h,0)-g(0,0)}{h}=\lim _{h\rightarrow 0}\frac{f(|h|)-f(0)}{h}=0$$

$$g_y(0,0)=\lim _{k\rightarrow 0} \frac{g(0,k)-g(0,0)}{k}=\lim _{k\to 0}\frac{f(|k|)-f(0)}{k}=0$$

Since, $f'(0)=0$ we would have $$\lim_{(h,k)\to (0,0)} \frac{g(h,k)-g(0,0))}{\sqrt{h^2+k^2}}=\lim _{t\to 0} \frac{f(t)-f(0)}{t}=f'(0)=0$$ Thus, $$\lim_{(h,k)\to (0,0)}\frac{g(h,k)-g(0,0)-\big( g_x(0,0)h+ g_y(0,0)k\big)}{\sqrt{h^2+k^2}}\\=\lim_{(h,k)\to (0,0)} \frac{g(h,k)-g(0,0)}{\sqrt{h^2+k^2}}=0$$

i.e., $g(x,y)$ is differentiable on $\mathbb{R}^2$.