Differentiation is about small changes. We interpret the differential $df$ as the “small change in $f$”, so that the basic differentiation operation becomes \begin{equation} f \longmapsto df \end{equation} which we describe as “zapping $f$ with $d$”.

The derivative rules from first term calculus can be written in differential form as: \begin{align} d(u^n) &= nu^{n-1}\,du \\ d(e^u) &= e^u\,du \\ d(\sin u) &= \cos u\,du \\ d(uv) &= u\,dv + v\,du \\ d(u+cv) &= du + c\,dv \end{align} where $c$ is a constant. The first three rules suffice to differentiate all elementary functions, 1) and the latter two, referred to as the Leibniz property and linearity, respectively, are distinguishing characteristics of many different notions of differentiation. The quotient rule is not needed, as it follows from the product rule together with the power rule. Derivatives of inverse functions are most easily computed directly, e.g. \begin{equation} u = \ln v \Longrightarrow v = e^u \Longrightarrow dv = e^u\,du = v\,du \Longrightarrow du = \frac{dv}{v} \end{equation}

Finally, the chain rule is conspicuous by its absence. For example, if $f=q^2$ and $q=\sin u$, then \begin{equation} df = 2q\,dq = 2\sin u\cos u\,du \end{equation} so that \begin{equation} \frac{df}{du} = \frac{df}{dq} \frac{dq}{du} \end{equation} so that the traditional chain rule is built into differential notation. Put differently, we have \begin{equation} df = \frac{df}{dq} \,dq = \frac{df}{du} \,du \end{equation}

The differential $df$ is not itself a derivative, as we have not yet specified “with respect to what”. Rather, the derivative of $f$ with respect to, say, $u$ is just the ratio of the small changes $df$ and $du$.

Functions of several variables can be differentiated without further ado. For example, suppose that $f=\sin u$ with $u=pq$. Using the product rule, we obtain \begin{equation} df = \cos u\,du = \cos u (p\,dq+q\,dp) = \cos(pq) (p\,dq+q\,dp) \end{equation}

It is important to realize that differentials are not themselves the answers to any physical questions. So what was the question? Perhaps the goal was to compute the derivative of $f$ with respect to $p$. Strictly speaking, this question is poorly posed, and one should seek the derivative of $f$ with respect to $p$ with $q$ held constant. But that's easy: If $q$ is constant, $dq=0$, and the desired derivative is just the coefficient of $dp$, written \begin{equation} \frac{\partial f}{\partial p} = q\,\cos(pq) \end{equation}

We have expressed $df$ in terms of $dp$ and $dq$; this is very much like finding the components of a vector (“$df$”) in terms of a basis (“$\{dp,dq\}$”). Furthermore, the “components” are just partial derivatives of $f$: \begin{equation} df = \frac{\partial f}{\partial p} \,dp + \frac{\partial f}{\partial q} \,dq \end{equation}

This interplay between calculus and algebra is at the heart of the study of differential forms.