Page 107 - Foundations Of Differential Calculus

90 5. On the Differentiation of Algebraic Functions of One Variable
By multiplication we obtain
2 3
qdp = αB dx +2αCx dx +3αDx dx +4αEx dx + ···
2 3
+βBx dx +2βCx dx +3βDx dx + ···
2 3
+ γBx dx +2γCx dx + ···
3
+ δBx dx + ··· ;
2 3
pdq = βA dx + βBx dx + βCx dx + βDx dx + ···
2 3
+2γAx dx +2γBx dx +2γCx dx + ···
2 3
+3δAx dx +3δBx dx + ···
3
+4 Ax dx + ··· .
From these we obtain the desired differential dy, which is the quotient
whose numerator is equal to

2
(αB − βA) dx +(2αC − 2γA) xdx +(3αD + βC − γB − 3δA) x dx
3
+(4αE +2βD − 2δB − 4 A) x dx
4
+(5αF +3βE + γD − δC − 3 B − 5ζA) x dx

and whose denominator is equal to

2 3 4 5 2
α + βx + γx + δx + x + ζx + ··· .
This expression is most accommodated to the expeditious differentiation
of any rational function. Since the numerator of the differential is made
up from coefficients of the numerator and denominator functions, it can be
obtained by inspection. The denominator of the differential is the square
of the denominator of the given function.

166. If in the given quotient either the numerator or the denominator, or
both, is made up of a product, then when the multiplication is performed
we have a form we have already differentiated. However, we give special
rules to make it easier to cover these cases.
Suppose the given quotient has the form y = pr/q.Welet pr = P. Then

dP = pdr + r dp.
Since y = P/q,wehave
qdP − Pdq
dy = ,
q 2
and after substituting for P and dP, we have the following result: