# Why is any arbitrary directional derivative always recoverable from the gradient?

I understand what partial derivatives, directional derivatives, and the gradient are. I can even follow symbolically from their definitions why:

$$D_\mathbf{u} f = \nabla f \cdot \mathbf{u}$$

But nonetheless I find it surprising that knowing the derivative in just 3 directions at a point (the gradient) is sufficient to figure out what it must be in any direction. The equation states it must be the case that the derivative in an arbitrary direction must be a weighted combination (dot product) the gradient with the weights determined by how x-axis'y and how y-axis'y the direction u is. What prevents me from constructing a function where this isn't true? Why can't a simultaneous increase in x and y give a dramatically different result than either alone? (e.g. a function that rises in the x+ direction and the y+ direction, but falls dramatically along the diagonal?)

• You might find this question and this one relevant. – amd Feb 9 '18 at 21:40
• Comments are not for extended discussion; this conversation has been moved to chat. – quid Feb 10 '18 at 12:35

Why can't a simultaneous increase in x and y give a dramatically different result than either alone? (e.g. a function that rises in the x+ direction and the y+ direction, but falls dramatically along the diagonal?)

Well, it can, but then the function won't be differentiable. One concrete example of a function that has different behavior in the $x$ and $y$ axes then it has in between is the function $z = r\sin(2 \theta)$, in cylindrical coordinates. This function is not differentiable at the origin. It is continuous at the origin and has slopes of $0$ in the $x$ and $y$ directions there - the $x$ and $y$ axes are both contained in the graph of the function. But in other directions the slopes at the origin can be anything else between $1$ and $-1$.

Remember that a point and two slopes in non-parallel directions are all that we need to completely determine a plane. So, if the tangent plane to the graph of $f(x,y)$ is well defined at a point, the slopes of the tangent plane in the $x^+$ and $y^+$ directions completely characterize the plane. A plane, being flat, can't increase along the $x^+$ and $y^+$ axes and decrease in between.

If a function tried to do that, it would not be differentiable at the point in question - it would not be well approximated by the plane that the gradient determines. This is the source of the definition of differentiability: a differentiable function has its slope in each direction determined by that direction and the slopes in the $x^+$ and $y^+$ directions.

The same thing happens in one dimension, we just get too used to it to see it. You might ask, "why does the behavior of a function in the $x^+$ direction determine the behavior in the $x^-$ direction? Why can't a function rise in both the $x^+$ and $x^-$ directions?". Of course, a function can do that, like $y = |x|$ does. But then the function will not be differentiable at the point in question, because it will not be well approximated by the line that is determined by the rate of change in the $x^+$ direction.

The situation in two or more variables is no different. In one dimension, the slope in the $x^+$ direction determines a line. In two dimensions, the slopes in the $x^+$ and $y^+$ directions determine a plane. In either case, we define the function to be differentiable if, around the point we started with, the function is well approximated by that line or plane in every direction that we can go, given the number of dimensions we are working with.

As I said in the comments, this is just the chain rule. Here's the formal argument.

By definition, the directional derivative of $f:\mathbb{R}^n\supset U\to\mathbb{R}$ at $p$ in the direction of $u$ is

$$D_uf(p):=\lim_{t\to0}\frac{f(p+tu)-f(p)}{t}$$

But this is also the definition of derivative of $g(t):=f(p+tu)$ at $t=0$. In other words,

$$D_uf(p)=g'(0)$$

Now, $g$ is a composite function: $g=f\circ\ell$, where

$$\ell(t)=p+tu$$

By the multivariate chain rule (I'm assuming the hypotheses of that theorem are satisfied),

$$g'(0)=(f\circ\ell)'(0)=f'\big(\ell(0)\big)\ell'(0)$$

But $\ell(0)=p$, $\ell'(0)=u$, and $f'(p)=\nabla f(p)$.

We conclude

$$D_uf(p)=g'(0)=\nabla f(p)u$$

(You don't need the dot product in this expression, if you interpret $\nabla f$ correctly as a row vector.)

As for the intuitive idea, the point is that in the infinitesimal limit (which is what derivatives measure), a differentiable function is well-approximated by a linear one, so its rate of change in any direction must be a linear function of the changes in the coordinate directions.

This is because of the definition of the gradient and in general of what it means for a function $f$ to be differentiable (at a point $p_0$ say).

Intuitively, because $f$ is differentiable at $p_0$ is aproximated by a linear function (which in the case of $\mathbb{R^n}$ is exactly the gradient)aproximates really well. So it is logical to believe that in a small neighbourhoud around $p_0$ $f$ behaves almost like a linear function. These aproximations are made more expliciti with the Taylor theorem.

Let me say that this way of thinking, while not rigorous,provides a great deal of intuition behind very importan theorems e.g. the inverse function theorem.

• Something is off in the first sentence of the second paragraph. – Carl Mummert Feb 9 '18 at 21:37

Geometric explanation

It is because for a differentiable function $f:\Bbb R\to\Bbb R^n$ an amount of $n$ vectors is enough to know the tangent plane. All tangent vectors lie in the tangent plane and an $n$-dimensional plane is uniquely determined by $n$ (linearly independent) vectors. If you have a function with a "sharp bend" (e.g. think about a roof), where all directional derivatives exist, but no unique tangent plane, then this will not work because not all tangent vectors lie in a plane.

Think about a differentiable function as a function which has unique tangent planes everywhere. Your counterexamples are not differentiable then.