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...?