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.

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