Page 102 - Foundations Of Differential Calculus

5. On the Differentiation of Algebraic Functions of One Variable 85
x, p becomes p + dp, and q becomes q + dq. The product pq is transformed
into
(p + dp)(q + dq)= pq + pdq + qdp + dp dq.

Hence, the differential of the product pq is equal to pdq + qdp + dp dq.
Since pdq and qdp are infinitely small of the first order, while dp dq is of
the second order, the last term vanishes, with the result that

d.pq = pdq + q dp.
It follows that the differential of the product pq consists of two members,
each of which is one factor multiplied by the differential of the other. From
this we easily deduce the differential of the triple product pqr.Ifwelet
qr = z, then pqr = pz and d.pqr = pdz + zdp. Since z = qr,wehave
dz = qdr + rdq, and after substituting for z and dz,wehave

d.pqr = pq dr + pr dq + qr dp.
In a similar way, if the quantity to be differentiated is a fourfold product,
then we have
d.pqrs = pqr ds + pqs dr + prs dq + qrs dp.

From this it should be easily seen what the differential of a product of many
factors will be.

I. If y =(a + x)(b − x), then

dy = −dx (a + x)+ dx (b − x)= −adx + bdx − 2x dx.
This same differential can be found by expanding the expression to
2
y = ab − ax + bx − x , so that by the previous rule,
dy = −adx + bdx − 2x dx.

1
2
2
II. If y = a − x ,welet
x
1
2
2
= p and a − x = q,
x
but since
−dx −xdx
dp = and dq = √ ,
2
x 2 a − x 2
we have
−dx dx
2
2
dy = pdq + qdp = √ − 2 a − x .
2
a − x 2 x