Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 5

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Lovelace also tried her hand at proving standard results in calculus, such as the general power rule, namely that $\frac{d}{dx} (x^n )=nx^{n-1}.$

The method she used, which was correct, was to find the difference quotient $\frac{(x+\theta)^n-x^n}{\theta},$ then use the binomial theorem to expand $(x+\theta)^n$, cancel the appropriate terms, and take the limit as $\theta \rightarrow 0$. ‘It strikes me,’ she told De Morgan, ‘as having the advantage in simplicity, & in referring to fewer requisite previous Propositions’ [LB 170, 10 Jan. [1841], f. 82v].

Her proof was perfectly sound, but there was one small problem—its reliance on the binomial theorem. She had learnt that theorem from De Morgan’s Elements of Algebra, in which the generally accepted proof of the time was given. Unfortunately, that proof relied on a lemma which stated that: $\lim_{w\rightarrow v}⁡\frac{v^n-w^n}{v-w}= nv^{n-1}.$ As De Morgan was quick to point out, ‘if you take the common proof of the binomial theorem, you are reasoning in a circle, for that proof requires that it should be shown that $⁡\frac{v^n-w^n}{v-w}$ has the limit $nv^{n-1}$ as $w$ approaches $v$. This is precisely the proposition which you have deduced from the binomial theorem’ [LB 170, [Jan. 1841], f. 34r]. A few days later, Lovelace acknowledged that ‘my proof of the limit for the function $x^n$ is a piece of circular argument’ [LB 170, 17 Jan. [1841], f. 85v]. As she explained: ‘It had not struck me that, calling $(x+\theta)=v$, the form $\frac{(x+\theta)^n-x^n}{\theta}$ becomes $\frac{v^n-x^n}{v-x}$ ’ [LB 170, [Jan. 1841], f. 91r].

As an exercise, instead of using Lovelace’s method, try proving the general power rule the way De Morgan did it (for positive integer values of $n$). He used a result equivalent to Theorem 2 from Peacock’s Collection of Examples (see Figure 13) [De Morgan 1836-42, 51]:

If $u$ be the product of n functions $PQR\ldots$ then the product of all but $P$ is $u/P$, and so on; whence we have $\frac{du}{dx}=\frac{u}{P}\frac{dP}{dx}+\frac{u}{Q} \frac{dQ}{dx}+\frac{u}{R} \frac{dR}{dx}+ \ldots$

You can check your proof here.

Adrian Rice (Randolph-Macon College), "Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Problem 5," Convergence (September 2021)