Given that the directional derivative is defined formally as:

$$ \nabla_\vec{v}\, f\left(\vec{x}\right) = \lim_{h \to 0} \frac{f\left(\vec{x} + h\vec{v}\right) - f\left(\vec{x}\right)}{h|\vec{v}|} $$

It's not exactly clear why the magnitude should matter here if we are taking some step $h\vec{v}$ if $h \to 0$. If the directional derivative can be calculated via $\nabla f\left(\vec{x}\right) \cdot \vec{v}$, any scalar to $\vec{v}$ will arbitrarily scale the magnitude of the directional derivative - but if we take the limit definition, shouldn't magnitude not matter as we are investigating how $f$ changes with some infinitesimal movement in the direction of $\vec{v}$?

One explanation was to imagine a function $f(x, y)$ representing the altitude via a surface, with a person moving around it. If a person were to move with double the velocity in a certain direction, they should have double to amount in change of height. But I can't reconcile this with the formal definition: why should doubling the velocity change how much $f$ changes if you move infinitesimally in the direction of the velocity?


Generally speaking, derivatives are rates of change, that is, they quantify the change in a function’s value per unit step. For a scalar-valued function $f$, a directional derivative measures how steep the graph of $f$ is in the direction of $\mathbf v$. To use your analogy, the steepness of the slope doesn’t depend on how fast you’re going when you measure it.

This is what your difference quotient for the directional computes. Compare it to the difference quotient for an ordinary derivative: $${f(x+\Delta x)-f(x)\over \Delta x}.$$ Geometrically, we evaluate $f$ at some distance $\Delta x$ from $x$ and divide by this distance to get a rate of change. The same thing is going on in the difference quotient for the directional derivative, with the additional information of the direction in which we’re looking, encoded in the fixed vector $\mathbf v$. The displacement from $\mathbf x$ is $h\mathbf v$, therefore to get a rate of change we have to divide by $\lVert h\mathbf v\rVert = h\lVert\mathbf v\rVert$. (It’s not $\lvert h\rvert\lVert\mathbf v\rVert$ for the same reasons that we don’t use $\lvert\Delta x\rvert$ in the ordinary derivative.) Some sources require $\mathbf v$ to be a unit vector, in which case the factor of $\lVert\mathbf v\rVert$ doesn’t appear explicitly in the difference quotient, but it’s lurking there nonetheless.

Incidentally, this limit definition is more fundamental than the expression $\nabla f\cdot\mathbf v$ since directional derivatives can exist where $f$ is not differentiable, or even where its partial derivatives don’t exist. Note, too, that partial derivatives are just a special case of directional derivative: $${f(\dots,x_i+h,\dots)-f(\dots,x_i,\dots) \over h} = {f(\mathbf x+h\mathbf e_i)-f(\mathbf x)\over h\lVert\mathbf e_i\rVert}.$$

It can be useful to define a directional derivative so that the length of $\mathbf v$ matters, but in that case it computes a linear approximation to the change in $f$ instead of a rate of change. Going back to your analogy, if you double the velocity, you double the change in altitude. The former definition, that of a rate of change, is much more common.

  • $\begingroup$ Okay, so there are two definitions: one explicitly about the (per unit) rate of change of $f$ versus some infinitesimal movement $\mathrm d \vec{v}$, and the other just about approximating $f$ linearly and seeing how it changes when moving with "velocity" $\vec{v}$? So these two definitions really have no equivalence unless we have a unit vector in $\nabla f \cdot \hat{u}$? $\endgroup$ – Andrew Li Apr 21 at 20:52
  • $\begingroup$ @AndrewLi Right, although they’re clearly related. For a unit vector $\mathbf u$, $f(\mathbf x+h\mathbf u) = f(\mathbf x)+hD_{\mathbf u}f(\mathbf x)+o(h)$: you get the linear approximation to the absolute change by multiplying the unit change rate by the distance. $\endgroup$ – amd Apr 21 at 20:55
  • $\begingroup$ Sorry for all the questions, but one last query: how would I compute the actual $\mathrm df / \mathrm d\vec{v}$ from the limit definition? Would it be $\nabla f \cdot \mathbf u$ since that would be looking at a change in $f$ per unit vector $\mathbf u$? $\endgroup$ – Andrew Li Apr 21 at 21:02
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    $\begingroup$ @AndrewLi If $f$ is differentiable, then $D_{\mathbf u}f=\nabla f\cdot\mathbf u$ (the proof is fairly straightforward), otherwise you usually have to fall back on the limit definition. $\endgroup$ – amd Apr 21 at 21:04
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    $\begingroup$ @AndrewLi Don’t apologize for all the questions. It shows that you’re thinking about the material and trying to get a deeper understanding of it. Kudos! $\endgroup$ – amd Apr 21 at 21:07

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