Differentiability of a two variable function $f(x,y)=\dfrac{1}{1+x-y}$ 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 ?
 A: If all partial derivatives of a function (over all possible variables) are continuous at some point, then the function is differentiable at that point.
A: 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 $y$.
A: 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 $$\lim\limits_{(h,k) \to (0,0)} \frac{f(h,k)-f(0,0)+h-k}{\sqrt{h^2+k^2}}=0$$
