# Differentiability of multivariable functions

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## Main Question or Discussion Point

What does it mean for a $f(x,y)$ to be differentiable at $(a,b)$? Do I have to somehow show $f(x,y)-f(a,b)-\nabla f(a,b)\cdot \left( x-a,y-b \right) =0$? To show the function is not though, it's enough to show, using the limit definition, that the partial derivative approaching in one direction is not equal to the partial from another direction right?

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andrewkirk
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Yes that is sufficient, but it is not necessary. A function may have both those partial derivatives exist and be equal, yet not be differentiable at that point.

Yes that is sufficient, but it is not necessary. A function may have both those partial derivatives exist and be equal, yet not be differentiable at that point.
So to summarize, because $f(x,y)$ is a surface, to show that the function is diff at a point, we have to take into account all the partial derivatives in all directions at the point, and they all have to exist. Another option is to only look at the partial with respect to x or y, and then approach in different directions; if they are different, then the function is already not diff. However, if they are the same, it still does not guarantee diff because we can approach in infinitely many ways? Then that means currently, I can only disprove whether or not a function is diff. Is this where the "gradient" definition comes in?

andrewkirk
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There is usually an easy way to prove differentiability, using the multidimensional differentiability theorem. All you need to do is show that both the $x$-direction and $y$-direction partial derivatives of $f$ exist and are continuous on an open set containing $(a,b)$

All you need to do is show that both the xxx-direction and yyy-direction partial derivatives of fff exist and are continuous on an open set containing (a,b)
What about all the other directions? I thought a tangent plane had to exist in order for $f(x,y)$ to be diff

andrewkirk
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The tangent plane will exist if the criteria for that theorem are met. Consider the function $f:\mathbb R^2\to\mathbb R$ whose value is 0 except when $x=y\neq 0$, in which case the value is 1.

$\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ both exist but $f$ does not satisfy the requirement that it be partial differentiable in the $x$ and $y$ directions throughout some open set containing (0,0), because any such set will contain a point $(x,x)$ at which neither the $x$ nor $y$ direction partial derivative exists.

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I don't understand, how do those partials exist? Here is a chart I saw in a video (arrow means "a" implies "b") Continuous partials as in a ${ C }^{ 1 }$ function. ${ C }^{ 1 }$ meaning only with respect to x and y, or all directions?

This function isn't ${ C }^{ 1 }$, but is differentiable right? Meaning it implies continuity and existence of partial derivatives Last edited:
andrewkirk
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I don't understand, how do those partials exist?
They exist because a partial derivative is defined as a limit, and it exists if the limit exists. Your understanding of this will be greatly aided if you try to work out the value of those two partial derivatives at (0,0) using the limit. If you don't remember the formula for the limit, you will find it in my primer on partial differentiation.

Stephen Tashi
Stephen Tashi
In general, there is a difference (both in definition and in numerical value) between a "double limit" $\lim_{(x,y) \rightarrow (ab)} g(x,y)$ and the two "iterated limits" $\lim_{x \rightarrow a} ( \lim_{y\rightarrow b} g(x,y))$ and $\lim_{y \rightarrow b} ( \lim_{x\rightarrow a} g(x,y))$. And the two iterated limits are not necessarily equal.