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Section 6.13 Interpreting Differentials

Using differentials allows algebraic operations to yield information about differentiation. Not only do we know that

\begin{equation} df = \Partial{f}{x}\,dx + \Partial{f}{y}\,dy\tag{6.13.1} \end{equation}

but we can run this argument in reverse.

Suppose we know that

\begin{equation} du = A\,dv + B\,dw .\tag{6.13.2} \end{equation}

where \(A\) and \(B\) may be complicated algebraic expressions involving \(u\text{,}\) \(v\text{,}\) and \(w\text{.}\) Then, by comparing Equation (6.13.1) and Equation (6.13.2), we see that

\begin{align*} A \amp= \Partial{u}{v} ,\\ B \amp= \Partial{u}{w} . \end{align*}

We call this process "interpreting the partial derivatives."

Furthermore, we can use algebra to solve for \(dv\text{,}\) obtaining

\begin{equation} dv = \frac{1}{A}\,du - \frac{B}{A}\,dw ,\tag{6.13.3} \end{equation}

and we can conclude that

\begin{align*} \left(\Partial{v}{u}\right)_w \amp= \frac{1}{A} ,\\ \left(\Partial{v}{w}\right)_u \amp= -\frac{B}{A} . \end{align*}

With so many variables in use at the same time, it became important to specify which ones are being held constant when taking derivatives; this is done by writing \(\left(\Partial{v}{u}\right)_w\) to denote the partial derivative of \(v\) with respect to \(u\) "with \(w\) held constant." We have therefore shown that

\begin{align*} \left(\Partial{u}{v}\right)_w \left(\Partial{v}{u}\right)_w \amp= 1 ,\\ \left(\Partial{u}{v}\right)_w \left(\Partial{v}{w}\right)_u + \left(\Partial{u}{w}\right)_v \amp= 0 , \end{align*}

and the latter of these equations is usually rewritten in the form

\begin{equation} \left(\Partial{u}{v}\right)_w \left(\Partial{v}{w}\right)_u \left(\Partial{w}{u}\right)_v = -1\tag{6.13.4} \end{equation}

and called the cyclic chain rule.

In practice, using algebra to rearrange equations involving differentials automatically incorporates the chain rule in all of these forms. It is often easier to rearrange and interpret than to use the formulas.