Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 4

Adrian Rice (Randolph-Macon College)

Unfortunately, Lovelace’s work for Problem 4 no longer exists, but she probably attempted to solve it by a repeated application of the product rule. While there is nothing inherently wrong with this approach, it is clear that Peacock intended the problem to be solved using one of three theorems he stated in a footnote on the same page (see Figure 13). These theorems are simply three cases of what we nowadays call logarithmic differentiation, which can indeed simplify the process of differentiating products of multiple functions. But Lovelace was none the wiser, being completely mystified by the content of Peacock’s note: ‘Not one of the three theorems it contains is intelligible to me,’ she wrote [LB 170, 10 Nov. [1840], f. 63v].

From page 2 of Peacock's calculus textbook.

Figure 13. Footnote on page 2 of George Peacock’s Collection of Examples of the Application of the Differential and Integral Calculus (1820).

If she had understood the meaning of these theorems, she could have applied theorem (2) to \(u=x^2  (a+x)^3  (b-x)^4\), where \(v_1=x^2\), \(v_2= (a+x)^3\), and \(v_3= (b-x)^4\). This would have given her
\[\begin{array}{ccl}du&=&x^2  (a+x)^3  (b-x)^4 \left \{\frac{2x}{x^2} + 3\frac{(a+x)^2}{(a+x)^3} - \frac{4(b-x)^3}{ (b-x)^4} \right \}dx \\
&=&x^2  (a+x)^3  (b-x)^4 \left \{\frac{2}{x}+\frac{3}{a+x}-\frac{4}{b-x}\right \}dx\\
&=&x^2  (a+x)^3  (b-x)^4 \left \{\frac{(2(a+x)(b-x)+3x(b-x)-4x(a+x)}{ x(a+x)(b-x)}\right \}dx\\
&=&x^2  (a+x)^3  (b-x)^4 \left \{\frac{2ab-6ax+5bx- 9x^2}{x(a+x)(b-x)}\right \}dx\\
&=&x (a+x)^2  (b-x)^3 (2ab-(6a-5b)x-9x^2 )dx.\end{array}\]
Thus Peacock’s answer was in fact correct.

Lovelace’s mistake was that she did not apply the chain rule to  \((b-x)^4\) correctly, and obtained \(4 (b-x)^3 \)instead of \(-4 (b-x)^3\). As De Morgan explained to her: ‘It is very common to suppose that if \(\phi(x)\) differentiated gives \(\psi(x)\), then \(\phi(-x)\) gives \(\psi(-x)\), but this should be \(\psi(-x)\times \mbox{diff. co.}(-x)\) or \(\psi(-x)\times -1\)’ [LB 170, 14 Nov. 1840, f. 20r]. Sure enough, once Lovelace rectified her mistake she obtained the correct answer.

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