Check differentiability of $f(x,y)=y\sin\frac{1}{x}$ at $(0,0)$

Question

Given , $f(x,y)=y\sin\frac{1}{x}$ when $x\ne0$ and $f(0,0)=0$ . Investigate differentiability at $(0,0)$ .

I've found that it is continuous at $(0,0)$ and the partial derivatives $f_x=0$ $\forall (x,y)$ and $f_y=\begin{cases} \sin\frac1x & \text{ if } x\ne0\\ 1 & \text{ if } x= 0 \end{cases}.$

For differentiability it is sufficient to show that both partial derivatives exist and one of them(confused between "both continuous" Or "one of them") is continuous about some neighbourhood of $(0,0)$ .

Now, I'm stuck. Please help how to think.

EDIT

$f(0,y)=y$

https://math.stackexchange.com/questions/853215/sufficient-conditions-for-differentiability-of-multivariate-functions see here — Learning, Feb 05 '18 at 13:00

GNUSupporter 8964民主女神地下教會 · Accepted Answer · 2024-08-29T20:18:23.633

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(Edited in response to Did and PrithiviRaj's comments.)

$$ \begin{aligned} f_x (0,0) &= \lim_{h \to 0} \frac{f(h,0)-f(0,0)}{h} \\ &= \lim_{h \to 0} \frac{0\sin\frac1h - 0}{h} = 0 \end{aligned} \\ \begin{aligned} f_y (0,0) &= \lim_{k \to 0} \frac{f(0,k)-f(0,0)}{k} \\ &= \lim_{k \to 0} \frac{k - 0}{k} = 1 \end{aligned} $$

In the definition of differentiability, approach the point $(0,0)$ via the curve $y = x^2$ to see that the limit $$ \lim_{(h,k) \to (0,0)} \frac{f(h,k) - f(0,0) - 0 \cdot h - 1 \cdot k}{\sqrt{h^2+k^2}} \tag{*} \label1$$ can't be defined. Take $k = h$.

\begin{equation} \frac{f(h,h)-f(0,0)-h}{\sqrt{h^2+h^2}} = \frac{h (\sin \frac{1}{h} - 1)}{\sqrt{h^2+h^2}} = \frac{\sin\frac{1}{h} - 1}{\sqrt2} \tag{$h\ne0$} \end{equation}

Observe that the denominator $\sqrt2$ is a constant, but the limit of $\sin(1/h)$ as $h \to 0$ is undefined.

To see this,

take $h_n = 1/(n\pi)$ so that $h_n\to0$ and $\sin(1/h_n) = 0$; but
take $h'_n = 1/((2n+1/2)\pi)$ so that $h'_n\to0$ and $\sin(1/h'_n) = 1$.

Hence \eqref{1} is undefined, and $f$ is not differentiable at $(0,0)$.

edited Aug 29 '24 at 20:18

answered Feb 05 '18 at 10:32

GNUSupporter 8964民主女神地下教會

17,998

but , it is sufficient not necessary , if $f_y$ and $f_x$ are not continuous , then $ f $ can be differentiable. – Feb 05 '18 at 12:01
@Martin Thanks for pointing my mistakes out. Sorry for that. I'm rewriting my answers. – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 13:00
Could not make any sense for your answer ! Explain it ? @GNUSupporter – Learning Feb 05 '18 at 13:08
@PrithiviRaj I've chosen the curve $y = x^2$ by experience. The logic is that (1) Differentiability implies existence of directional derivatives (2) Directional derivatives can be defined by a parametrization along a curve. (3) My arg: there exists some curve at which the directional derivative can't be defined, so it's not differentiable. You may want to see the wiki page of directional derivatives (Notation and Definition). – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 13:20
"Directional derivatives" are limits along straight lines, not along other curves. – Did Feb 05 '18 at 13:23
@Did Thanks your for comment. I would like to draw your attention to the example "directional derivative along a curve" in a lecture notes from Harvard. – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 13:26
@GNUSupporter , Is this true ,if directional derivatives do not exist , then it ($f$)can't be differentiable ? – Learning Feb 05 '18 at 14:27
@PrithiviRaj If $f$ is differentiable at $x$, then all directional derivatives at $x$ exists. I'm showing its contrapositive form. – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 14:30
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okay! , thanks! , now it is clear, nice approach. – Learning Feb 05 '18 at 14:31
@GNUSupporter Seems quite idiosyncratic to me. Interestingly (and not very surprisingly), the wiki page you linked to, stays on the usual practice of calling directional derivatives, derivatives in a... direction. – Did Feb 05 '18 at 15:58
@Did In the Notation part, it's said the vector "v is a parameterization of a curve". – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 16:02
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Indeed, and since v is constant, the "curve" one obtains is... a straight line. – Did Feb 05 '18 at 17:04
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@Did Thanks for explaining this. If I understand you correctly, your point is the formula $$\nabla_{\bf v} f({\bf x}) = \nabla f({\bf x}) \cdot {\bf v}$$ for differentiable $f$ at $x$ takes constant ${\bf v}$. I've edited my answer so as to use the definition of differentiability. – GNUSupporter 8964民主女神地下教會 Feb 05 '18 at 18:53
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One way to define differentiability is that the derivative along any differentiable curve is equal to the dot product of the gradient and the tangent vector along the curve. (In this way, the $\mathbf{v}$ need not be constant, and in fact it is important, if this condition is to be sufficient, that the curve need not be a line, as the example of $y^3/x$ shows.) So along $y=x^2$, you could take $\lim_{h\to0}(f(h,h^2)-f(0,0))/h = \lim_{h\to0}h\sin(1/h)$, which is not $(0,1)\cdot(1,0)=0$ (in fact the limit is undefined). – Toby Bartels Nov 13 '18 at 23:22
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How does the denominator tend to 1 when h tends to 0 – Upstart Aug 28 '24 at 18:19
@Upstart thx for pointing that out. hope it's correct now – GNUSupporter 8964民主女神地下教會 Aug 29 '24 at 20:19

Check differentiability of $f(x,y)=y\sin\frac{1}{x}$ at $(0,0)$

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