Is $$f(x,y) =\begin{cases} x^2+2x+2y & \text{ for } (x,y)\neq (0,0) \\ y^2 & \text{ for } (x,y)=(0,0) \end{cases}$$ derivable? But its partials are not continuous?

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    $\begingroup$ The function you have defined is the same as the polynomial $f(x,y)=x^2+2x+2y$, so it is differentiable everywhere in $R^2$ and has continuous partial derivatives everywhere also. $\endgroup$
    – user84413
    Aug 27, 2014 at 18:17
  • $\begingroup$ It's not even true in one dimension that a differentiable function $f$ must have $f'$ continuous. (Take $f(x) = x^2 \sin 1/x$, for example.) Are you alluding to the result that the existence and continuity of partial derivatives implies differentiability? $\endgroup$
    – anomaly
    Aug 27, 2014 at 19:30
  • $\begingroup$ No the opposite differentiability don't imply that the parcials derivatives are continuous however they exist @anomaly $\endgroup$
    – studentNk
    Aug 27, 2014 at 19:34
  • $\begingroup$ Is that true? @anomaly $\endgroup$
    – studentNk
    Aug 27, 2014 at 19:36
  • $\begingroup$ Right: A differentiable (as opposed to $C^1$) function may have partial derivatives that are not continuous. In one dimension, take $f(x) = x^2 \sin 1/x$; in multiple dimensions, take something like $r^2 \sin 1/r$ for $r^2 = x_1^2 + \cdots + x_n^2$. $\endgroup$
    – anomaly
    Aug 27, 2014 at 19:43

1 Answer 1


If a function $f(x,y)$ is continuous at a point in its domain and its partial derivatives exist at that point, then $f(x,y)$ is said to be derivable at that point.

Here, $f(x,y)$ is continuous at $(0,0)$.

And if both $f^\prime_x$ and $f^\prime_y$ exist at $(0,0)$, then $f(x,y)$ is derivable at $(0,0)$.

So, $f^\prime_x =\lim_{h \to 0}$ $\frac{f(h,0)-f(0,0)}{h}$

Hence, $f^\prime_x = 2$

Similarly, $f^\prime_y =\lim_{h \to 0}$ $\frac{f(0,h)-f(0,0)}{h}$

Hence, $f^\prime_y = 2$

So, $f(x,y)$ is derivable at $(0,0)$.

And at all points in $R^2$,except $(0,0)$, it's pretty obvious that $f(x,y)$ is derivable as $f(x,y)$ is simply a polynomial in that domain.

Hence, $f(x,y)$ is derivable on $R^2$.

  • $\begingroup$ But when (x,y)=(0,0) f(x,y)=y^2 and fy(x,y)=2y hence fy(0,0)=0 $\neq$ 2 and f`x(0,0)=0 $\neq$2 @user3679857 $\endgroup$
    – studentNk
    Aug 27, 2014 at 20:16
  • $\begingroup$ You can't write $f^\prime_y = 2y$ before actually proving that $f(x,y)$ is derivable on that $(0,0)$. $\endgroup$ Aug 27, 2014 at 20:23
  • $\begingroup$ First you prove $f(x,y)$ is derivable and then you use the results based on differentiation. $\endgroup$ Aug 27, 2014 at 20:26
  • $\begingroup$ If its not derivable then its parcials don't exist ?@user3679857 $\endgroup$
    – studentNk
    Aug 27, 2014 at 20:32
  • $\begingroup$ The fact that the partial derivetives exist does not suffice to say that the function it's differentiable there. They have to be continuos as well $\endgroup$
    – Ant
    Aug 28, 2014 at 8:01

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