$f(x,y) = g(\sqrt{x^{2}+y^{2}})$ Prove that f is differentiable at $(0,0)$ iff $g'(0)=0$

Question

$f(x,y) = g(\sqrt{x^{2}+y^{2}})$. Prove that f is differentiable at $(0,0)$ iff $g'(0)=0$

This was a question on my midterm a few days ago. I've been thinking about it for a while and still cannot come up with an answer.

My guess would be to use partials and prove that they're continuous at $(0,0)$ but I don't know how to incorporate the fact that $g'(0)=0$.

Answer 1

Note that $f(x,0)=g(|x|)$ and $f(0,y)=g(|y|)$ so

$$\frac{\partial f}{\partial x}(0,0)=\frac{\partial f }{\partial y}(0,0)=g^{\prime}(0)$$

Now the condition for differentiability is $$\lim\limits_{(h,k)\rightarrow 0} \frac{f(h,k)-f(0,0)-g^{\prime}(0)h-g^{\prime}(0)k}{\sqrt{h^2+k^2}} =0 $$

Now $$\lim\limits_{(h,k)\rightarrow 0} \frac{f(h,k)-f(0,0)}{\sqrt{h^2+k^2}} = \lim\limits_{r\rightarrow 0} \frac{g(r)-g(0)}{r} =g^{\prime}(0)$$

So the condition becomes

$$\lim\limits_{(h,k)\rightarrow 0} g^{\prime}(0)\left(1-\frac{h+k}{\sqrt{h^2+k^2}}\right) =0 $$ Note however that

$\lim\limits_{(h,k)\rightarrow 0} \frac{h+k}{\sqrt{h^2+k^2}} $ does not exist, so we must have $g^{\prime}(0)=0$. Conversely, $g^{\prime}(0)=0$ gives the limit $0$ and shows that $f$ is differentiable.

Note that the calculation of the values if the partials, $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ is irrelevant to the solution to the problem. We were just to say $\frac{\partial f}{\partial x}(0,0)=a$ and $\frac{\partial f}{\partial y}(0,0)=b$ the differentiability condition becomes $$\lim\limits_{(h,k)\rightarrow 0} \frac{f(h,k)-f(0,0)-ah-bk}{\sqrt{h^2+k^2}} =0 $$ And the nonexistence of the limits $\lim\limits_{(h,k)\rightarrow 0} \frac{h}{\sqrt{h^2+k^2}} $ and $\lim\limits_{(h,k)\rightarrow 0} \frac{h}{\sqrt{h^2+k^2}} $ gives $g^{\prime}(0)=a=b=0$.

Answer 2

Set $t=\sqrt{x^2+y^2}$ then we have $f=g(t)$ then $$f_x=g_t.(t_x)=g_t.(\dfrac{x}{\sqrt{x^2+y^2}})$$ Notice that we must have $g_t=0$ otherwise we can not set $f_x$ when $x,y=0$.

Answer 3

First show by taking $y = x$ that if $f$ is differentiable at $(0, 0)$, then $g$ is differentiable at $0$. The chain rule then gives \begin{align*}\frac{\partial f}{\partial x} &= \frac{x}{\sqrt{x^2 + y^2}} g'(\sqrt{x^2 + y^2})\\\frac{\partial f}{\partial y} &= \frac{y}{\sqrt{x^2 + y^2}} g'(\sqrt{x^2 + y^2}) \end{align*} away from $(x, y) = (0, 0)$. By taking $y = \pm x$ again, show that the partial derivatives are well-defined at $(0, 0)$ only if $g'(0) = 0$. In the opposite direction, show that the partial derivatives above are well-defined and continuous at $(0, 0)$ if $g'(0) = 0$, thus forcing $f$ to be differentiable there.

Answer 4

Another approach: We have, $$ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} g'(\sqrt{x^2 + y^2}) $$ $$ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} g'(\sqrt{x^2 + y^2}) $$ So, for $f$ to be differentiable at (0,0), we need, $lim_{(x,y)->(0,0)} \left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)$ to exist. Setting, $x=r cos(\theta)$ and $y=rsin(\theta)$, $$ \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} = cos(\theta)g'(rcos(\theta))+sin(\theta)g'(rsin(\theta))$$. Noting that as you approach (0,0) from different directions, $\theta$ takes different values, this expression can be well defined, only if $g'(0) = 0$