Question about continuous differentiability

Question

Consider the function $f(x,y) = \frac{x}{1+\sqrt{x^2+y^2}}$.

Its derivative with respect to $x$ can be calculated to be $\frac{1 + \frac{y^2}{\sqrt{x^2 + y^2}}}{1 + x^2 + y^2 + 2 \sqrt{x^2 + y^2}}$.

Is it correct to say that $\frac{\partial f(x,y)} {\partial x}$ is continuous? I ask because it seems that if both $x$ and $y$ are zero, then the derivative is undefined.

But $f(x,y)$ itself is undefined at $(0,0)$. Yes; ultimately, you calculate $f_x$ and $f_y$ , and see if they are continuous. — user99680, Dec 01 '13 at 21:43
@user99680 Sorry I wrote the wrong function. I have corrected. — Sunny88, Dec 01 '13 at 21:52
Actually, the partial with respect to $x$ is defined at $(0,0)$ now. — user99680, Dec 01 '13 at 21:54
but notice you have the $1$'s that you are adding to the zeros. Do the computation using $(x,y)=(0,0)$. — user99680, Dec 01 '13 at 23:04
@user99680 In the numerator you have the term $1+\frac{y^2}{\sqrt{x^2+y^2}}$. If you substitute x and y by zeros, u get $1 + \frac{0}{0}$ which is undefined. — Sunny88, Dec 02 '13 at 06:46
You can see that $0\leq y/\sqrt{x^2+y^2}\leq1$, so it cannot diverge. — Vincenzo Tibullo, Dec 02 '13 at 21:20

Vincenzo Tibullo · Accepted Answer · 2013-12-03T07:41:15.603

1

The obtained formula for the derivative is undefined in $(0,0)$, here you can use the definition: $$ \frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{h}{1+\sqrt{h^2}}}{h}=1 $$ This corresponds to the limit in $(0,0)$ of the obtained formula, given that $$ \lim_{(x,y)\to(0,0)}\frac{y^2}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}y\cdot\frac{y}{\sqrt{x^2+y^2}}=0 $$ as the limit of the product of an infinitesimal function by a bounded function.

edited Dec 03 '13 at 07:41

answered Dec 02 '13 at 21:42

Vincenzo Tibullo

10,955

So, can we say that $f(x,y)$ is continuously differentiable with respect to $x$? – Sunny88 Dec 03 '13 at 05:26
@Sunny88: yes, it is – Vincenzo Tibullo Dec 03 '13 at 06:47
Thank you. This helped me to understand as well: http://www.freemathhelp.com/forum/threads/85005-Question-about-cotinuous-differentiability?p=349378&posted=1#post349378 – Sunny88 Dec 03 '13 at 18:51

Question about continuous differentiability

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