Difference between Directional Derivative x Chain Rule for Scalar Fields

Question

could you guys help me out with an issue I am having.

1. What's the difference between the "Directional Derivative" and "Chain Rule for Scalar Fields"? In meaning and the formulae?

I don't know if I got it right but both of them have the same formula:

$$ g'(\vec{r(t)}) = \nabla(g(t)) \cdot \vec{r'(t)} \ for \ the \ Chain \ Rule$$

And

$$ Derivative = \nabla (g(t)) \cdot \vec{r(t)} \ for \ the \ Directional\ Derivative$$

And to me they both seem to have the same meaning, since when we use the chain rule, we are using a vector whose direction is defined by the parameters.

Answer 1

You are taking the derivative (both cases) of a composition $\mathbb R^1\stackrel{r}\to\mathbb R^n\stackrel{g}\to\mathbb R^1$ defined by $g\circ r(t)=g(r(t)$ to get $g'(r(t))\cdot r'(t)$ or in classical notation $$\frac{dg\circ r}{dt}(t)=\nabla g(r(t))\cdot r'(t),$$ which is a particular instance of the general cases $\mathbb R^1\stackrel{r}\to\mathbb R^n\stackrel{G}\to\mathbb R^m$, that also comply a chain rule but, for $G'$ one has a matrix $m\times n$.

Answer 2

Your definition of directional derivatives is actually a consequence of the chain rule, but it only applies when $f$ is differentiable at $t$. However, it’s not the fundamental definition of a directional derivative, which can exist even when the function isn’t differentiable. For example, $f(x,y)=(x^2y)^{1/3}$ has a directional derivative in every direction at the origin, but is not differentiable there.

The fundamental definition of a directional derivative looks very much like the definition of an ordinary derivative: $$D_{\mathbf v}f(\mathbf x) = \lim_{h\to0}{f(\mathbf x+h\mathbf v) - f(\mathbf x)\over h}.$$ Note that the partial derivatives that you’re no doubt familiar with are actually special cases of directional derivatives. It’s a theorem that when $f$ is differentiable at $\mathbf x$, then $D_{\mathbf v}f(\mathbf x)=\nabla f(\mathbf x)\cdot\mathbf v$. Geometrically, this is because differentiability implies the existence of a tangent (hyper)plane to the graph, and the directional derivative is then just the slope of this plane in the direction of $\mathbf v$.