Differentials such as $df$ are rarely themselves the answer to any physical question. So what good is the Master Formula? The short answer is that you can use it to answer any question about how $f$ changes. Here are some examples.

1. Suppose you are an ant walking in a puddle on a flat table. The depth of the puddle is given by $h(x,y)$. You are given $x$ and $y$ as functions of time $t$. How fast is the depth of water through which you are walking changing per unit time?

This problem is asking for the derivative of $h$ with respect to $t$. So divide the Master Formula by $dt$ to get $$ {dh\over dt} = \grad{h} \cdot {d\rr\over dt} $$ where $\rr$ describes the particular path you are taking. The factor $d\rr\over dt$ is simply your velocity. This dot product is easy to evaluate, and yields the answer to the question.

(There are of course many ways to solve this problem; which method you choose may depend on how your path is described. It is often easiest to simply insert the given expressions for $x$ and $y$ in terms of $t$ directly into $h$, then differentiate the resulting function of a single variable, thus calculating the left-hand side directly.)

2. You are another ant on the same surface, moving on a path with $y=3x$. How fast is the depth changing compared with $x$?

This problem is asking for the derivative of $h$ with respect to $x$ as you move along the path; note that this is the total derivative $dh\over dx$, not the partial derivative $\Partial{h}{x}$, which would only be appropriate if $y$ were constant along the path. So divide the Master Formula by $dx$ to get $$ {dh\over dx} = \grad{h} \cdot {d\rr\over dx} $$ then use what you know ($y=3x$) to relate the changes in $x$ and $y$ ($dy=3\,dx$), so that $$ d\rr = dx\,\xhat+dy\,\yhat = dx\,\xhat+3\,dx\,\yhat = (\xhat+3\,\yhat) \,dx $$ to obtain $$ {d\rr\over dx} = \xhat + 3\,\yhat $$ Evaluating the dot product yields the answer to the question.

3. You are still moving on the same surface, but now the question is, how fast is the depth changing per unit distance along the path?

This problem is asking for the derivative of $h$ with respect to arclength $ds$. We can divide the Master Formula by $ds$, which leads to $$ {dh\over ds} = \grad{h} \cdot {d\rr\over ds} $$ Unfortunately, it is often difficult to determine $s$; it is not always possible to express $h$ as a function of $s$. On the other hand, all we need to know is that $$ ds = |d\rr| $$ so that dividing $d\rr$ by $ds$ is just dividing by its length; the result must be a unit vector! Which unit vector? The one tangent to your path, namely the unit tangent vector $\TT$, so \begin{equation} {dh\over ds} = \grad{h} \cdot \TT \label{Directional} \end{equation} Evaluating the dot product answers the question, without ever worrying about arclength.

We have just seen that the derivative of $f$ along a curve splits into two parts: a derivative of $f$ (namely $\grad{f}$), and a derivative of the curve ($d\rr/du$). But the latter depends only on the tangent direction of the curve at the given point, not on the detailed shape of the curve. This leads us to the concept of the directional derivative of $f$ at a particular point $\rr=\rr_0=\rr(u_0)$ along the vector $\vv$, which is traditionally defined as follows: 1) $$D_{\vv} f = \lim_{\epsilon\to0} {f(\rr_0+\epsilon\vv)-f(\rr_0) \over \epsilon}$$ According to the above discussion, this derivative is just $df/du$ along the tangent line, which according to the Master Formula is $$D_{\vv} f = \grad{f} \cdot \vv$$ In Example 3 above, the left-hand side of ($\ref{Directional}$) is just the directional derivative of $h$ in the direction $\TT$, and could have been denoted by $\DD{\TT}h$.