- #1
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First of all, I don't have a concrete example for this, but I hope it's not too hard to understand what I'm trying to get at.
For a multivariable function of, say, 2 variables x and y, the gradient at a point only depends on the existence of partial x and partial y, right? In other words, if these exist, then as far as I know the gradient exists. One of the definitions of the directional derivative gives it as grad(f) dot product with u, where u is the unit vector in the direction where you want to find the derivative of the function. But what if the function is not continuous in this direction at that point and therefore cannot have a derivative? Using the above definition of the directional derivative would be meaningless, I assume. But in most of the places I've seen that give this formula for the directional derivative, they don't mention that the function has to be continuous/differentiable in this direction for the derivative to exist. Maybe it's so obvious they don't feel like it needs to be mentioned?
Basically, what I'm getting at is, how can you know if a directional derivative exists in every direction at a certain point in the domain of a function? Is it simply a matter of the function being continuous in every direction from that point, and if so how can you be certain that a given function is continuous in every direction?
I guess one way to check for the existence of a directional derivative is to use the limit definition, right? I'm referring to the following:
lim (h->0) of [f(x + h*cos(theta), y + h*sin(theta)) - f(x, y)] / h. And if this limit exists, then the directional derivative exists, correct?
Is there any other (perhaps easier) method you can use to determine that a directional derivative exists in every direction at a particular point? I'm thinking it may be possible just by examining the equation of the function you're given but I'm not 100% certain.
For a multivariable function of, say, 2 variables x and y, the gradient at a point only depends on the existence of partial x and partial y, right? In other words, if these exist, then as far as I know the gradient exists. One of the definitions of the directional derivative gives it as grad(f) dot product with u, where u is the unit vector in the direction where you want to find the derivative of the function. But what if the function is not continuous in this direction at that point and therefore cannot have a derivative? Using the above definition of the directional derivative would be meaningless, I assume. But in most of the places I've seen that give this formula for the directional derivative, they don't mention that the function has to be continuous/differentiable in this direction for the derivative to exist. Maybe it's so obvious they don't feel like it needs to be mentioned?
Basically, what I'm getting at is, how can you know if a directional derivative exists in every direction at a certain point in the domain of a function? Is it simply a matter of the function being continuous in every direction from that point, and if so how can you be certain that a given function is continuous in every direction?
I guess one way to check for the existence of a directional derivative is to use the limit definition, right? I'm referring to the following:
lim (h->0) of [f(x + h*cos(theta), y + h*sin(theta)) - f(x, y)] / h. And if this limit exists, then the directional derivative exists, correct?
Is there any other (perhaps easier) method you can use to determine that a directional derivative exists in every direction at a particular point? I'm thinking it may be possible just by examining the equation of the function you're given but I'm not 100% certain.