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I have encountered this in an online PDE course I'm following but I've never really been exposed to it. I've looked for the 'formal' definitions but I've never really understood any concept by looking at the formal, mathematical definition so can anyone elucidate this concept?

ColinvanNoordt
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    Rate of change in any particular direction is defined by a directional derivative. So, it is a dot product of gradient vector(max. rate of change pointing vector) and the unit vector in the desired direction. – kaka Aug 07 '13 at 21:03

2 Answers2

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The derivative is just the rate of change of a function of one variable. Well, the directional derivative is the rate of change you get after converting a function of many variables into a function of one variable. You do this by picking a "direction" and traveling along that direction. As qaphla points out in a comment, what we call a "partial derivative" is exactly a directional derivative taken in a certain direction. So the partial with respect to $x$ is the directional in the $x$ direction, and so on.

For example if I want to take a directional derivative of $f(x, y) = xy$ at $(a, b)$ in the direction $u = \langle 1, 2\rangle$ then my new function is $$h(t) = f(a + t, b + 2t) = (a + t)(b + 2t) = ab + 2at + bt + 2t^2$$ and it's derivative is $h' = 2a + b + 4t$. In $(a + t, b + 2t)$ the point $(a, b)$ corresponds to $t = 0$ so $\partial_uf(a, b) = h'(0) = 2a + b$.

On the other hand, if I want to find the partial with respect to $x$ then I choose the $x$ direction $u = \langle 1, 0\rangle$. I get $h(t) = (a + t)b$, $h'(t) = b$, and so $\partial_xf(a, b) = b$. Notice this is exactly what I get if I just follow the normal rules for differentiating $f$ and treat $x$ as a variable and treat $y$ as if it were a number (and hence had derivative $0$).

Jim
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  • Hmm. I would have thought you convert a function of many variables into a function of one by taking its partial derivative with respect to that one variable.. however, how would one pick such a direction? Can you give a rudimentary example in your answer? – ColinvanNoordt Aug 07 '13 at 21:03
  • Taking a partial derivative with respect to a variable does not necessarily eliminate that variable. For instance $f(x, y) = x^2y + y^3$ has partial $\partial_xf = 2xy$, so it is still a function of two variables. – Jim Aug 07 '13 at 21:05
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    The partial derivatives are a special example of directional derivatives in the directions (1,0,...,0), (0,1,0,...,0),...,(0,0,...,0,1). In general, however, it is possible to take the derivative at a given point along any vector. (Though some people define directional derivatives only for vectors with norm 1.) – qaphla Aug 07 '13 at 21:06
  • Oops, I derped, you are correct. – ColinvanNoordt Aug 07 '13 at 21:06
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Suppose $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function. Then at a point $a\in\mathbb{R}$, the derivative $f'(a)$ tells you how quickly $f(x)$ increases (or decreases, if $f'(a)$ is negative) as $x$ increases, for $x$ near $a$. Now if we have a function $f:\mathbb{R}^n\to \mathbb{R}$, it doesn't make sense to talk about $x$ "increasing", but we can instead ask how $f(x)$ changes as $x$ moves in a prescribed direction $u$, for $x$ near a point $a\in \mathbb{R}^n$. This is what the directional derivative $\partial_uf(a)$ tells us.

bradhd
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  • +1 this is an interesting view. I guess I almost agree. If the function is continuously differentiable then the directional derivative tells us how the function is changing in some direction at a point. I would go so far as to say the function "increases in that direction" (or decreases)... but, sadly the example given http://math.stackexchange.com/questions/447104/continuous-function-with-linear-directional-derivatives-total-differentiability/447197#comment962924_447197 goes to show the perils of not assuming $C^1$ – James S. Cook Aug 08 '13 at 01:39