Differentiability of $g:=f(\sqrt{x^2+y^2})$ for a $C^1$ function with $f'(0)=0$ NBHM $2008$ Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.
Define for all $x,y\in \mathbb{R}$,
$$g(x,y)=f(\sqrt{x^2+y^2})$$
Pick out the true statements :


*

*$g$ is differentiable on $\mathbb{R}^2$

*$g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$

*$g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$


I do not have much idea of differentiation of multivariable calculus.
I just tried out some examples to get some idea..
I have to take some function whose derivative at $0$ has to be zero.
for simplicity I would take $f(x)=x^2$ Then I would get :

$g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$

This would suggest me that second option has more chances as $f(0)=0$ in this case.
I would now take $f(x)=x^2+1$ then :

$g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$

So, I would now eliminate second option and third option also...
So, It is now got confirmed that i have to work out to "Prove" that first option is correct.
Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.
I have to look for partial derivatives first :
$$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$
$$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$
I have used here that $f$ is differentiable on $\mathbb{R}$
As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :
$\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.

As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$
As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$

So we have continuous partial derivatives for $g(x,y)$..
So, I am willing to conclude that:

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

I would be thankful if some one can confirm if the answer correct.
I would be much more thankful if some one can let me know if there are any gaps in my argument.
This is just a proof verification Question..
Thank you. 
 A: Your formulas establish that $g$ is $C^1$ away from the origin, but establish nothing at the origin. (Your continuity claim for $x/\sqrt{x^2+y^2}$ certainly fails at the origin.) Check the definition and ask if there is a linear map $Dg(0)\colon \Bbb R^2\to \Bbb R$ with the appropriate properties.
EDIT: To wit, we need
$$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)-\big(\partial g/\partial x(0,0)h+\partial g/\partial y(0,0)k\big)}{\sqrt{h^2+k^2}}=0. \tag{$\star$}$$
You need to actually go back to the definition of partial derivative to compute the partial derivatives of $g$ at $(0,0)$. A sneaky alternative is to guess the answer (hmmm, $0$?) and use the limit ($\star$) to deduce that it must be correct.
A: Just for the sake of clarity and to keep every thing at one place and for my reference, I would like to sum it up :
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 first of all find partial derivatives : 
$$g_x(a,b)=\lim _{h\rightarrow 0} \frac{g(a+h,b)-g(a,b)}{h}$$
$$g_y(a,b)=\lim _{k\rightarrow 0} \frac{g(a,b+k)-g(a,b)}{k}$$
at $(0,0)$ we would then have 
$$g_x(0,0)=\lim _{h\rightarrow 0} \frac{g(h,0)-g(0,0)}{h}=\lim _{h\rightarrow 0}\frac{f(|h|)-f(0)}{h}=0$$
$$g_y(0,0)=\lim _{k\rightarrow 0} \frac{g(0,k)-g(0,0)}{k}=\lim _{k\rightarrow 0}\frac{f(|k|)-f(0)}{k}=0$$
So, we have $g_x(0,0)=0,g_y(0,0)=0$.. 
So we would then have 
$$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)-\big( g_x(0,0)h+ g_y(0,0)k\big)}{\sqrt{h^2+k^2}}=\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)}{\sqrt{h^2+k^2}}$$
As $f'(0)=0$ we would have $$\lim _{h\rightarrow 0} \frac{f(h)-f(0)}{h}=f'(0)=0$$ and in particular $$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0))}{\sqrt{h^2+k^2}}=0$$
So we would then have,
$$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)-\big( g_x(0,0)h+ g_y(0,0)k\big)}{\sqrt{h^2+k^2}}=\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)}{\sqrt{h^2+k^2}}=0$$
Thus, 
$$\lim_{h,k\to 0} \frac{g(h,k)-g(0,0)-\big( g_x(0,0)h+ g_y(0,0)k\big)}{\sqrt{h^2+k^2}}=0$$
i.e., $g(x,y)$ is differentiable on $\mathbb{R}^2$.
Done :)
P.S : I have checked every thing twice but then if you think there is some argument missing please let me know.. 
Thank you :)
