Question about cotinuous differentiability

For other readers, the poster's question is as follows:

\(\displaystyle \mbox{Consider the func}\mbox{tion }\, f(x,\, y)\, =\, \dfrac{x}{1\, +\, \sqrt{x^2\, +\, y^2}}\)

\(\displaystyle \mbox{Its deriv}\mbox{ative with respect to }\, x\, \mbox{ can be calc}\mbox{ulated to be :}\)

. . . . .\(\displaystyle \dfrac{1\, +\, \dfrac{y^2}{\sqrt{x^2\, +\, y^2}}}{1\, +\, x^2\, +\, y^2\, +\, 2\sqrt{x^2\, +\, y^2}}\)

\(\displaystyle \mbox{Is it correct to say that }\, \dfrac{\partial f(x,y)}{\partial x}\, \mbox{ is cont}\mbox{inuous?}\)

\(\displaystyle \mbox{I ask because it seems that if both }\, x\, \mbox{ and }\, y\, \mbox{ are zero,}\)

\(\displaystyle \mbox{then the deriv}\mbox{ative is undefined.}\)
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Your last statement, "If both x and y are zero, the derivative is undefined" is incorrect.
The partial derivative with respect to x is taken holding y constant. With y= 0, we have
\(\displaystyle f(x)= \frac{x}{1+ |x|}\). In fact, the question asked, whether the partial derivative with respect to x was continuous, wouldn't make much sense if the derivative did not even exist at (0, 0)!

For x= 0 and h> 0, the difference quotient is \(\displaystyle \frac{\frac{h}{1+ h}}{h}= \frac{1}{1+ h}\) which goes to 1 as h goes to 0. If h< 0, the difference quotient is \(\displaystyle \frac{1}{1- h}\) which still goes to 1 as h goes to 0. The partial derivative with respect to x, at (0, 0), is 1.
 
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