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Find the partial derivatives of $f (x,y) = x^2\sin(1/x)+y^2$ at $(0,0)$

Is this function differentiable at the origin? $$f(x,y) = x^2\sin(1/x)+y^2 \quad \text{if $(x,y) \ne 0$} $$ $$ = y^2 \qquad \text{if $(x,y) = 0$} $$

I found that $$\frac{df}{dx} =2x\sin(1/x)-\cos(1/x)$$ $$\frac{df}{dy} =y^2$$

df/dy(0,0) is easy it's df/dy=0^2=0

But df/dx(0,0) is a bit more complicated as $\sin(1/x)$ is undefined at x=0 perhaps i can use squeeze theorem to say that the limit of df/dx as it approaches x=0 is 0 but that would be the limit and not df/dx(0,0)

How do i find $\frac{df}{dx}(0,0)$

Hanul Jeon
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Patrick
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  • I would worry about that cosine term for $x$ going to zero, (Not that it matters much, but your derivative with respect to $y$ is not correct) – imranfat Oct 14 '15 at 01:50
  • $(x,y) = 0$ makes no sense at all. – zhw. Oct 14 '15 at 02:33
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    This question is not a duplicate. I'm not sure what people are thinking. – zhw. Oct 14 '15 at 04:50
  • I agree. Although the central difficulty is addressed in the other question, this one has complications that are not covered there, and therefore should not be considered a duplicate. – Paul Sinclair Oct 15 '15 at 13:08

2 Answers2

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Using that you have: $$ f(x,y) =\left\{ \begin{aligned} &x^2\sin(1/x)+y^2 \quad &\text{if} \quad &(x,y) \ne (0,0) \\ &y^2 \qquad &\text{if} \quad &x=y=0 \end{aligned}\right. $$

Applying the limit definition of derivative gives: $$ \begin{aligned} \frac{\partial f}{\partial x}(0,0)&=\lim_{h\rightarrow 0}\frac{f(h,0)-f(0,0)}{h}\\ &=\lim_{h\rightarrow 0}\frac{h^2 \sin(1/h)+0^2}{h}\\ &=\lim_{h\rightarrow 0}h \sin(1/h)\\ &=0. \end{aligned} $$

This limit is zero, thus $f_x(0,0)=0$.

jdods
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The correct problem is surely $f(x,y) = x^2\sin(1/x) + y^2, x \ne 0,$ $f(x,y) = y^2, x = 0.$ Then for $x\ne 0,$ $f(x,0) = x^2\sin(1/x).$ Hence $\partial f/\partial x (0,0) = 0.$ Since $f(0,y) = y^2,$ we easily see $\partial f/\partial y (0,0) = 0.$

As for differentiability: Define $g(x)$ on $\mathbb {R}$ by $g(x)= x^2\sin(1/x), x \ne 0, g(0) = 0.$ Then $g$ is differentiable on $\mathbb {R}.$ Hence $G(x,y) = g(x)$ is differentiable on $\mathbb {R}^2.$ Since $f(x,y) = G(x,y) + y^2,$ a sum of differentiable functions on $\mathbb {R}^2,$ $f(x,y)$ is differentiable everywhere on $\mathbb R ^2.$

zhw.
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