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Section 6.5 Rules for Differentials

We can eliminate \(y\) from (6.4.5) by writing

\begin{equation} d(x^2) = 2x\,dx ,\tag{6.5.1} \end{equation}

and we can go even further by replacing the ubiquitous variable \(x\) by any physical or geometric quantity, such as \(u\text{.}\) The beauty of this approach is that differentiation is easy once you have convinced yourself of a few basic rules.

Let's start with some simple functions. For instance, the power rule for derivatives says that

\begin{equation} \frac{d}{du}(u^n) = nu^{n-1}\tag{6.5.2} \end{equation}

which in differential notation becomes \(d(u^n)=nu^{n-1}\,du\text{.}\)


We use \(u\) and later \(v\) to denote any quantity. It might be the case that \(u=x\text{,}\) or that \(u=f(x)\text{,}\) or that \(u=f(x,y)\text{,}\) or that \(u\) depends on other quantities. it doesn't matter.


“Taking the differential” or “zapping with \(d\)” is an operation; \(d\) itself is an operator, that acts on functions.

Applying this construction to the derivatives of elementary functions, we obtain the basic differentiation formulas in differential form, namely:

\begin{align*} d\left(u^n\right) \amp = nu^{n-1} \,du ,\\ d\left(e^u\right) \amp = e^u \,du ,\\ d(\sin u) \amp = \cos u \,du ,\\ d(\cos u) \amp = -\sin u \,du ,\\ d(\ln u) \amp = \frac{1}{u} \> du ,\\ d(\tan u) \amp = \frac{1}{\cos^2u} \> du . \end{align*}

So how do we use these formulas to compute derivatives?