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I'm studying Mathematical Analysis II for a university course. There is a training exercise that asks me to:

Find the partial derivatives at $(1,0)$ of $f(x,y)$, where:

$f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ when $(x,y)\neq (0,0)$ and

$f(x,y) = 0$ when $(x,y)= (0,0)$

So far, I've used the definition as per the other examples. For example, let's start with $f_x(a,b)=g'(a)=\lim_{h\to 0}\frac{f(a+h,b) - f(a,b)}{h}$

Applied to my problem, I've got $f_x(1,0)=\lim_{h\to 0}\frac{f(1+h,0) - f(1,0)}{h}$

And this is where I stall. According to other examples, this is supposed to equal $\frac{0}{h}=0$ , but HOW? And then I'm supposed to get the limit to +inf?

Am I using the wrong methodology?

Community
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Dimitris S.
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3 Answers3

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Given the function $f$ defined as

$$f(x,y)= \begin{cases} \frac{xy}{\sqrt{x^2+y^2}}&,(x,y)\ne 0\\\\ 0&,(x,y)=(0,0) \end{cases}$$

we have

$$\frac{\partial f(x,0)}{\partial x}=\lim_{x\to 0}\frac{f(x+h,0)-f(x.0)}{h}=\lim_{h\to 0}\frac{\frac{(x+h)(0)}{\sqrt{(x+h)^2+o^2}}-\frac{(x)(0)}{\sqrt{(x)^2+o^2}}}{h}=0$$

Note that while both partial derivatives $f_x(0,0)=0$, $f_y(0,0)=1$, the Derivative of $f$ does not exist at the origin. We can see this by using the definition of the derivative

$$\lim_{h,k\to 0}\frac{f(0+h,0+k)-f(0,0)-f_x(0,0)h-f_y(0,0)k}{\sqrt{h^2+k^2}}=\lim_{h,k\to 0}\frac{\frac{hk}{\sqrt{h^2+k^2}}-0-0(h)-(1)k}{\sqrt{h^2+k^2}}$$

Since this limit does not exist, $f$ is not differentiable at the origin.

Mark Viola
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    Thank you for the solution. One small thing: as per skyking's analytical answer, $f_y$ comes out 1, but as per your way, $f_y$ yields zero. What gives? Shouldn't both ways yield the same? – Dimitris S. Sep 04 '15 at 21:53
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    +Dr.MV Correction, I think $f_y$ indeed yields 1, not zero. – Dimitris S. Sep 04 '15 at 22:06
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    It looks like it's quite different thing that was calculated. First of all the function isn't even the same, second the partial derivates are not calculated at the same point. No wonder the result come out different. – skyking Sep 04 '15 at 22:08
  • @DimitrisS. Thanks for the catch. I was working from a different definition of $f$. I have edited accordingly. – Mark Viola Sep 05 '15 at 01:34
  • @skyking Thanks for the catch. A +1 for the useful comment. I was working from a different definition of $f$ for some reason (bad eyes maybe??). Anyway, because of your useful comment, I have edited accordingly. – Mark Viola Sep 05 '15 at 01:35
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$f(1+h,0)=\frac{1+h. 0}{\sqrt{(1+h)^2+0^2}}=0$ similarly $f(1,0)=0$ thus $f_x(1,0)=\lim_{h\to 0}\frac{f(1+h,0) - f(1,0)}{h}=\lim_{h\to 0}\frac{0}{h}=0$

R.N
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At this level you should be allowed to use results from Analysis I (use the tools you already have). Which means that you can calculate the derivatives directly. For example partial derivate of $f(x,y)$ with respect to x at a point $(1,0)$ is the derivate of the function $\phi(x) = f(x,0)$ at the point $x=0$. So we can derive it as if the variable we're not deriving with respect to was a constant:

${\partial\over\partial x}xy(x^2+y^2)^{-1/2} = y(x^2+y^2)^{-1/2} - {1\over2}xy(x^2+y^2)^{-3/2}2x$ ${\partial\over\partial y}xy(x^2+y^2)^{-1/2} = x(x^2+y^2)^{-1/2} - {1\over2}xy(x^2+y^2)^{-3/2}2y$

as for the value at $(0,0)$ it would have no influence for the derivatives at $(1,0)$. Setting $x=1$ and $y=0$ yields:

$f_x'(1,0) = 0(1^2+0^2)^{-1/2} - {1\over2}1\cdot0(1^2+0^2)^{-3/2}2\cdot1 = 0$

$f_y'(1,0) = 1(1^2+0^2)^{-1/2} - {1\over2}1\cdot0(1^2+0^2)^{-3/2}2\cdot0 = 1$

skyking
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  • Thank you for providing the analytical way! It helps a lot. Please check my comment on Dr. MV's answer above, also. – Dimitris S. Sep 04 '15 at 21:54