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We're given the following function :

$$f(x,y)=\dfrac{1}{1+x-y}$$

Now , how to prove that the given function is differentiable at $(0,0)$ ?

I found out the partial derivatives as $f_x(0,0)=(-1)$ and $f_y(0,0)=1$ ,

Clearly the partial derivatives are continuous , but that doesn't guarantee differentiability , does it ?

Is there any other way to prove the same ?

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    $\begingroup$ "Clearly the partial derivatives are continuous , but that doesn't guarantee differentiability , does it ?" It does. "Is there any other way to prove the same ?" Yes, for instance using the definition. $\endgroup$
    – Git Gud
    Aug 22, 2015 at 10:58
  • $\begingroup$ See this math.stackexchange.com/questions/1007709/… $\endgroup$
    – user87543
    Aug 22, 2015 at 11:00
  • $\begingroup$ $f(x,y) = 1$ for x>0 , y>0 and $f(x,y) = 0$ ,otherwise. This function $f(x,y)$ is non-differentiable , even though the partial derivatives are continuous.. @GitGud $\endgroup$
    – User9523
    Aug 22, 2015 at 14:14
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    $\begingroup$ @RohitDuggal The partial derivatives are not defined for all points $(x,y)$ in an open neighbourhood of either $(x,0)$ or $(0,y)$ for $x\geq0$ and $y\geq0$ for the function you gave. However, in the case of the function you had in the question, we have that the partial derivatives are continuous in an open neighbourhood of $(0,0)$. This is what makes all the difference. $\endgroup$
    – Scounged
    Aug 22, 2015 at 15:37

3 Answers 3

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You have several ways to approach the question.

First one $f$ is the ratio of two differentiable functions, the denominator one not vanishing in the neighborhood of the origin. Hence $f$ is differentiable at the origin.

Second one Using a theorem stating that if $f$ is continuous in an open set $U$ and has continuous partial derivatives in $U$ then $f$ is continuously differentiable at all points in $U$.

Third one Using the definition of the derivative, prove that $$\lim\limits_{(h,k) \to (0,0)} \frac{f(h,k)-f(0,0)+h-k}{\sqrt{h^2+k^2}}=0$$

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If all partial derivatives of a function (over all possible variables) are continuous at some point, then the function is differentiable at that point.

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  • $\begingroup$ Dont we need this to be true in a neighborhood of a point? I cant think of an example where this is true but I think it is. $\endgroup$
    – MSIS
    May 25, 2020 at 23:50
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You need to show that the limit definition of the partial derivative from each direction is the same value. Hence

$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0^{-}} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0^{-}} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^{-}} -\frac{1}{(1+h)^{2}} = -1$ (Using l'hopital's rule).

$\frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0^{+}} \frac{f(0+h,0)-f(0,0)}{h} = \lim_{h \to 0^{-}} \frac{\frac{1}{1+h} - 1}{h} = \lim_{h \to 0^{-}} -\frac{1}{(1+h)^{2}} = -1$.

Therefore the function is differentiable with respect to $x$. Now rinse and repeat the same thing for $y$ to determine if $f$ is differentiable with respect to $y$.

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