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I've had some trouble with the (visual) intuition behind the directional derivatives so I decided to take a step back and look up the visual intuition behind partial derivatives, which I think I do understand. See picture below

enter image description here

As I understood it, we basically have the purple paraboloid (?) which is a function of (x,y) and then we have the gray plane which is the plane in the direction of the x-axis. If you intersect the 2 planes I would say you get a parabola. If you have a certain point specified on the paraboloid, you can find its partial derivative in the direction of x.

The way I draw the connection to a directional derivative is just by saying that you can tilt the gray plane in any direction and find the derivative. Is this correct, and if not, what's wrong?

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Yes, I believe your intuition is correct. –  A.E Aug 16 '13 at 20:00
    
Only thing to remember with directional derivatives is that different authors handle scaling differently. Depending on context, the directional derivative might or might not depend on the norm of the direction vector (where it does, we usually think of the directional derivative as the rate of change experienced by an observer passing through the point at a velocity equal to the direction vector). –  Jonathan Y. Aug 16 '13 at 21:00

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The "canonical" way of explaining a directional derivative is this. At the point of interest, draw the gradient vector (normal to the tangent hyperplane). A directional derivative is the projection of the gradient onto your chosen direction, i.e., the scalar product of the two.

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The answer is slightly misleading, as the existence of (some) directional derivatives is independent of the existence of all partial derivatives, i.e. a gradient vector. It is, however, accurate for differentiable functions. –  Jonathan Y. Aug 16 '13 at 21:13

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