Intuition behind directional derivative (geometrically)?

Question

The directional derivative in the direction <0, 1> is < ∂f/∂x, ∂f, ∂y> ⋅ <0, 1>. This makes perfect sense because this comes out to taking ∂f/∂x and multiplying it to 1, and adding it to (∂f/∂y)*(0), which is zero, coming to an end result of ∂f/∂x. I do not understand how this can be generalized. How can we say, for each and every possible function, that the rate of change at a given point in the direction will be a(∂f/∂x) + b(∂f/ ∂y)? Someone explained the directional derivative as being the sum of the vector's component in the x direction multiplied by a pure step in the x direction, and the vectors component in the y direction multiplied by a pure step in the y direction. For example, how can we say that for every possible function, the directional derivative in the direction <1/sqrt(2), 1/sqrt(2)> is equal to (1/sqrt(2))∂f/∂x + (1/sqrt(2))∂f/∂y? I would think things would vary from function to function. What relationship am I missing here? Please help!

Answer 1

Let $z=Ax+By+C$ be the tangent plane of the graph $z=f(x,y)$ at the point in question; thus $A=\partial f/\partial x$ and $B=\partial f/\partial y$ at that point. (This assumes that $f$ is differentiable, of course, but so does your question to begin with.)

Then the directional derivative with respect to the unit vector $(a,b)$ is just how much $z$ changes along the tangent plane if you add $a$ to $x$ and add $b$ to $y$. Which is obviously $aA+bB$.