Thanks David! Alright, if I understand your answer, the definition of a function being differentiable at a certain point is that it has a tangent plane. Once we identify a plane that agrees with two slopes at that point, that plane MUST be the tangent plane at that point, since the two slopes uniquely identify the plane. Okay, I accept that. However, at the end you say that there are other ways to show differentiability without having to check the slope...is there a way to connect those "ways" with the tangent plane "definition" intuitively? Thanks!

Hmm...but I can't see it even for very simple functions. In your answer, you say that we don't have to do a delta-epsilon proof in single-variable analysis. But, for example, take the function $f(x,y)=x^2$. If a tangent line agrees with the slope $2x$ at a certain point...well, there's only one slope to agree with, so I KNOW that the line will be tangent. But, now take the surface $f(x,y)=x^2$. A tangent plane at a certain point must have a slope of $2x$ in the $x$ direction and $0$ in the $y$ direction.

So tangent lines drawn on the plane in those two directions will be tangent to the surface. But, how can I intuitively see that tangent lines drawn on the plane in ANY direction will be tangent to the function at that point? Also, in my previous comment, I meant "take the function $f(x) = x^2$" the first time. Thanks again.

Wait--how do you know that $f(x) = x^2$ has only one slope at any given point? Either you looked at a graph that looks smooth, and you trust that there isn't some microscopic funny stuff going on, or you've done a delta-epsilon proof, or you know that the product of two differentiable function is differentiable, you know $x^2 = x \cdot x,$ and you know that $g(x) = x$ is differentiable. But there are really only two approaches, leap of faith or proof.

So you can look at a 3D graph of $f(x,y) = x^2,$ consider one of the horizontal lines on the graph (it has constant $x$ value) and convince yourself by observation that there's a plane that is tangent to the surface all along that line (maybe the same way you persuade yourself there is a tangent line to $g(x) = x^2$). And maybe you can convince yourself there are no funny bits of the surface popping out at an angle from the plane at the point of tangency. I wouldn't consider that a proof, but it might be intuitive.

On further thought, look at Lee Mosher's answer. He points out that there is actually a theorem about the very questions you're asking. So the definition alone is useful, but it does not answer everything.

Okay, let me try again. As someone pointed out in the comments, that the tangent line didn't cross the function only applied if the function is convex (or concave) like $x^2$. Lets try making a tangent line to $f(x)=x^2$. If the tangent line agrees with the slope in the one direction it can agree with the slope (the $x$ direction), I know that the tangent line won't cross the parabola. However, for the graph of $f(x,y)=x^2$, although I know lines on the plane won't cross the function when moving in the $x$ direction or in the $y$ direction...

... since the plane agrees with the slope of the surface in those two directions, how do I know that lines on the plane won't cross the surface if we draw them in another direction?

By "crossing the surface" you mean "touches the surface at an angle", right? Because the tangent to $f(x) = x^3$ crosses the curve at $x=0,$ and the line $y=0$ touches the graph of $g(x) = \lvert x\rvert$ without crossing it, but $x^3$ is differentiable at $0$ whereas $\lvert x\rvert$ is not.

Hey David. Yep, that's what I meant...but, lets consider just convex functions then...look at my update to the question! How can I tell that lines on the plane in some other direction (besides the $x$ and $y$ directions) won't cross the surface?