[16r] My dear Lady Lovelace

You have taken a proper time to begin with
Incommensurables and if the subject interests you, I should recommend
you to continue.  You understand of course that your Diff^l 
Calculus must be delayed from time to time while you make up
those points of Algebra and Trigonometry which you have
left behind.
D.C. p. 53.  As in page 22 refers to the method of proving that
if \(P=2Q\), lim. of \(P=2.\) lim. of \(Q\) 
In similar way it may be shown that if

 \(\frac{\Delta u}{\Delta x}\cdot\frac{\Delta x}{\Delta u}=1\)  lim.\ of \(\frac{\Delta u}{\Delta x}\times\) lim.\ of \(\frac{\Delta x}{\Delta u}=1\) 

 With reference to your remark remember that 

 \(\frac{\Delta u}{\Delta x}\cdot\frac{\Delta x}{\Delta u}=1\) and  \(\frac{a}{b}\times\frac{b}{a}=1\)   are the same proposition

 But \(\frac{du}{dx}\times\frac{dx}{du}=1\) and \(\frac{a}{b}\times\frac{b}{a}=1\)   are not the same

\( \frac{\Delta u}{\Delta x}\times\frac{\Delta x}{\Delta u}=1\)   by common algebra  \(\frac{a}{b}\times\frac{b}{a}=\frac{ab}{ab}=1\) 

But we cannot say \(\frac{du}{dx}\times\frac{dx}{du}=\frac{du\ dx}{dx\ du}=1\) 
because \(\frac{du}{dx}\) is a mere symbol to denote limit of \(\frac{\Delta u}{\Delta x}\) and \(du\) and \(dx\) 
have no separate meaning

[16v] N. & M. p. 17

 The erratum exists ['but the misprint is' crossed out] and must

 be set right as you propose

 ['for \(\frac{q_1}{p_1}-\frac{q_2}{p_2}\) ' crossed out]

 The lengthiness of the proof arises from the necessity
of adapting a very common algebraical theory to Euclid's
method.

You should try some of the examples of differentiation
in Peacock's book.  Remember that there are some
misprints in it.  You will not have to go through 
it to try a little of everything.

When the article Proportion appears in the Penny Cycl.
which it will in a few weeks, I recommend your
attention to it

With remembrances to Lord Lovelace I am

 Yours truly

  ADeMorgan

69 Gower St.

 Sunday M^g Sept^r 27/40