Section 5.5 Rules for Differentials
We can eliminate \(y\) from (5.4.5) by writing
\begin{equation}
d(x^2) = 2x\,dx ,\tag{5.5.1}
\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{5.5.2}
\end{equation}
which in differential notation becomes \(d(u^n)=nu^{n-1}\,du\text{.}\)
Notation.
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.
Usage.
“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?
