7
$\begingroup$

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 ?

$\endgroup$
  • 2
    $\begingroup$ "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 '15 at 10:58
  • $\begingroup$ See this math.stackexchange.com/questions/1007709/… $\endgroup$ – user87543 Aug 22 '15 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 '15 at 14:14
  • $\begingroup$ @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 '15 at 15:37
13
$\begingroup$

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 $$\lim\limits_{(h,k) \to (0,0)} \frac{f(h,k)-f(0,0)+h-k}{\sqrt{h^2+k^2}}=0$$

$\endgroup$
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.