What was Newton's proof of the Product Rule and why was it "not satisfactory"?

Question

Piermont (1928) :

As is well known, the proof that Newton gave in his Principia that $(uv)' = uv'+vu'$ is not satisfactory.

What was Newton's proof of the Product Rule and why was it "not satisfactory" ?

Answer 1

While Leibniz gave Proof 1 , Newton gave Proof 2.

Summarized Proof 1 :

Consider the rectangle of sides $u$ & $v$. Area is $A=uv$. Increment both to get new Area $(u+du)(v+dv)$. Expand that & ignore $du \times dv$ which is "infinitely" smaller than the other "infinitesimals"
Hence , we get the Profuct rule $d(uv)=udv+vdu$

Summarized Proof 2 :

Consider the Area of the rectangle $A=uv$.
Increment both by "half infinitesimals" , $du/2$ & $dv/2$ to get $A+da/2$.
Decrement both by "half infinitesimals" , $du/2$ & $dv/2$ to get $A-da/2$.
Subtract the two to get $da/2+da/2=da$ , which will give the Product rule.

While Proof 1 works in general , Proof 2 works only when we take "half infinitesimals" : It is a great coincidence !! If we have incremented by $1/3$ & decremented by $2/3$ , it will not work.

Thus Proof 2 is not really a Proof : though it gives a good intuition , it is not rigorous & hence it is "not satisfactory" here.

reference :
https://math.libretexts.org/Bookshelves/Analysis/Real_Analysis_(Boman_and_Rogers)/02%3A_Calculus_in_the_17th_and_18th_Centuries/2.01%3A_Newton_and_Leibniz_Get_Started

Answer 2

This answer just adds the original text and translation.

Principia (3e, 1726, p. 244):

Cas. 1. Rectangulum quodvis motu perpetuo auctum $AB$, ubi de lateribus $A$ & $B$ deerant momentorum dimidia $\frac 1 2 a$ & $\frac 1 2 b$, fuit $A - \frac 1 2 a$ in $B - \frac 1 2 b$, seu $AB - \frac 1 2 aB - \frac 1 2 bA + \frac 1 4 ab$; & quam primum latera $A$ & $B$ alteris momentorum dimidiis aucta sunt, evadit $A + \frac 1 2 a$ in $B + \frac 1 2 b$ seu $AB + \frac 1 2 aB + \frac 1 2 bA + \frac 1 4 ab$. De hoc rectangulo subducatur rectangulum prius, & manebit excessus $aB + bA$. Igitur laterum incrementis totis $a$ & $b$ generatur rectanguli incrementum $aB + bA$. Q.E.D.

Translation (1999):

CASE 1. Any rectangle $AB$ increased by continual motion, when the halves of the moments, $\frac 1 2 a$ and $\frac 1 2 b$, were lacking from the sides $A$ and $B$, was $A - \frac 1 2 a$ multiplied by $B - \frac 1 2 b$, or $AB - \frac 1 2 aB - \frac 1 2 bA + \frac 1 4 ab$, and as soon as the sides $A$ and $B$ have been increased by the other halves of the moments, it comes out $A + \frac 1 2 a$ multiplied by $B + \frac 1 2 b$, or $AB + \frac 1 2 aB + \frac 1 2 bA + \frac 1 4 ab$. Subtract the former rectangle from this rectangle, and there will remain the excess $aB + bA$. Therefore by the total increments $a$ and $b$ of the sides there is generated the increment $aB$ + $bA$ of the rectangle. Q.E.D.