Let $f(x,y)=\sqrt{x^2+(y-1)^2}$. Study the differentiability of the function at the point $(0,1).$

I know that the derivative of a multi variable function is calculated as follows:

$$\lim_{h\to0}\frac {\|f(x+h,y+h)-f(x,y)-J(h)\|_{\mathbb R}}{\|h\|_{\mathbb{R}^2}}$$

How do I actually use this on this function?

  • $\begingroup$ That's not the definition. What you wrote doesn't make sense. Is $h$ a real number or an element of $\mathbb{R}^2$? $\endgroup$ – José Carlos Santos Jan 21 at 18:29
  • $\begingroup$ It's on the wikipedia page ......... en.wikipedia.org/wiki/… $\endgroup$ – C. Cristi Jan 21 at 18:30
  • $\begingroup$ My guess is, it's an element of $\mathbb R^2$ $\endgroup$ – C. Cristi Jan 21 at 18:31
  • $\begingroup$ If it's an element of $\mathbb{R}^2$, then what's the meaning of $x+h$? $\endgroup$ – José Carlos Santos Jan 21 at 18:34
  • $\begingroup$ @JoséCarlosSantos You're right, I know, but I have no idea, I just want to study the differentiability and that's the definition I found on that site $\endgroup$ – C. Cristi Jan 21 at 18:35

Check the partial derivatives at $\;(0,1)\;$:

$$\begin{cases}f'_x=\cfrac x{\sqrt{x^2+(y-1)^2}}\implies f'_x(0,1)=...\text{doesn't exist}\\{}\\ f'(y)=\frac {y-1}{\sqrt{x^2+(y-1)^2}}\implies f'_y(0,1)=...\text{ doesn't exist}\end{cases}$$

and thus $\;f\;$ cannot be differentiable at $\;(0,1)\;$

  • $\begingroup$ What if one partial derivative exists and the other does not exist? $\endgroup$ – C. Cristi Jan 21 at 18:56
  • $\begingroup$ The function still isn't differentiable at that point. $\endgroup$ – DonAntonio Jan 21 at 19:08

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