Directional derivative expression

We have $n=\sqrt{{\mathbf N}\cdot{\mathbf N}}$, where ${\mathbf N}$ is the normal vector to a curve, let's accept ${\mathbf N}=\ddot{{\mathbf r}}$, say the curve is unit-speed. We also have a scalar function which depends only on $n$, namely $f= f(n)$. Is it true that the directional derivative

${\mathbf N}\cdot \nabla f(n) = n \frac{df(n)}{dn}$ ?

For comparison, the same case but with the usual gradient is

$\nabla f(r)= \frac{{\mathbf r}}{r}\frac{df(r)}{dr}$, which means that ${\mathbf r}\cdot \nabla f(r)= r \frac{df(r)}{dr}$, where $r=\sqrt{{\mathbf r}\cdot {\mathbf r}}$

Then ${\mathbf N}\cdot \nabla f(n)= n \frac{df(n)}{dn}$ is also true...?

• I don't think this makes much sense. The OP defines the quantity $a$ as the length of some vector $A$. Hence $a$ is a constant. But next it is used as the variable in some function. I don't see any relation between what the LHS (contraction of $A$ with the gradient) represents and what the RHS does. – M. Wind Jun 1 '15 at 2:04
• fixed the issue... – Jay Jun 1 '15 at 2:13
• I think what you wrote before is also true, in fact, modulo a little argument about the chain-rule and an agreement about notation, it is the definition of the directional derivative in the $A$-direction. We can write $\partial f /\partial a = (\nabla f) \cdot \hat{a}$ where $\hat{a} = A/\sqrt{A \cdot A}$. – James S. Cook Jun 1 '15 at 2:15
• The real question is what you mean by $d/dn$. How would you define this in terms of a limiting process on functions ? – James S. Cook Jun 1 '15 at 2:19
• $\frac{d}{dn}$ is the derivative w/respect to the length of vector ${\mathbf N}$. Say we have a scalar-valued function $f = f(n) = f(|\ddot{{\mathbf r}}|)$. Then ${\mathbf N}\cdot \nabla f = n\frac{df}{dn}$ – Jay Jun 1 '15 at 2:24