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For ease of notation and understanding, lets just stick to a function $f:\mathbb{R}^2\to\mathbb{R}$. Let $(x_0,y_0)$ be some point in $\mathbb{R}^2$. Suppose that the function is differentiable at $(x_0,y_0)$. My first question:

Does differentiability imply the existence of a number $D_f(x_0,y_0)$ such that $$\lim_{(x,y)\to(x_0,y_0)}\dfrac{f(x,y)-f(x_0,y_0)}{||(x,y)-(x_0,y_0)||}=D_f(x_0,y_0)$$

To me, the above limit seems to be the most straightforward and most intuitive definition of the derivative, because it really gives you an idea of the 'rate of change of the function'. At the same time, I can't ignore the following definition based on linear approximations:

A function $f:\mathbb{R}^2\to\mathbb{R}$ is said to be differentiable at $(x_0,y_0)$ if both partials $\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ exist and a function $\epsilon(h,k)$ exists such that $$f(x_0+h,y_0+k)-f(x_0,y_0)=\dfrac{\partial f}{\partial x}(x_0,y_0)h+\dfrac{\partial f}{\partial y}(x_0,y_0)+\epsilon(h,k)\sqrt{h^2+k^2}$$ such that $\epsilon(h,k)\to 0$ as $\sqrt{h^2+k^2}\to 0$

The second definition seems like a generalisation, while the first one seems to be the true one.

Now finally, my doubt is that if differentiability really means the first definition, then won't the partials be same valued at $(x_0,y_0)$? If not, what is differentiability then? How does it stick to its original 'change' concept?

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  • $\begingroup$ Have you tried your definition on the function $f(x,y) = x$? Does it produce results that you expect? $\endgroup$
    – user251257
    Apr 9 '17 at 20:50
  • $\begingroup$ The rate of change depends of the direction. Your intuition is wrong. $\endgroup$ Apr 10 '17 at 19:44
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The problem with your proposed definition is that it doesn't even work in $\mathbb R^1.$ Functions like $f(x)=x$ would be nowhere differentiable for example.

I would also note that the linear appoximation definition you gave is a little redundant in that there is no need to assume the partial derivatives exist. That they exist falls right out of the usual definition here.

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  • $\begingroup$ wait, why will the definition won't work for $f(x)=x$? The limit comes out to be $1$, which is true! $\endgroup$
    – codetalker
    Apr 14 '17 at 17:11
  • $\begingroup$ @Siddhant No, remember in your proposed redefinition, the denominator in the difference quotient is sign insensitive; it is always positive. So if $f(x) = x,$ then $$\lim_{x\to a} \frac{f(x)-f(a)}{|x-a|}$$ fails to exist for every $a.$ $\endgroup$
    – zhw.
    Apr 14 '17 at 17:16
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Take $(x_0,y_0) = (0,0)$, $f(x,y) = ax + by$. The limit $$ \lim_{(x,y)\to(x_0,y_0)}\frac{f(x,y)-f(x_0,y_0)}{\|(x,y)-(x_0,y_0)\|} = \lim_{(x,y)\to(0,0)}\frac{ax + by}{\|(x,y)\|} $$ does not exists (except in the trivial case $a = b = 0$) because the ratio depends of the direction: if $x = r\cos\theta$, $y = r\sin\theta$, $$\lim_{r\to 0}\frac{ar\cos\theta + br\sin\theta}r = a\cos\theta + b\sin\theta.$$

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