Calculus: Analytic Geometry and Calculus, with Vectors

156 Functions, limits, derivatives

Therefore, an application of a theorem on limits gives

TX =

C

TX +c,TX*

As has been remarked, putting c = ci = 1 gives (3.561) and putting

ci = 0 gives (3.562).

Postponing (3.563) and (3.564), we start proving the product formula

(3.565) by setting y = uv. Then

Y + AY = (u + Au) (v + Av)

= uv + u Av + v Au + Au Av,

so Ay = u AV + v Au + Au Av. Dividing by Ax and inserting an extra

factor Ax in the numerator and denominator of the last term give

(3.57) - =uAv+vAu+Au AVAx.

Ax Ax Ax Ax Ax

Taking limits as Ax approaches zero gives

dx=uax+Vdx+dxdx°'

The last term is zero, and this proves (3.565). Proof of the quotient

formula (3.566) is very similar, but the formula is important and we shall

prove it. Let y = u/v. Then

(3.571)

Y+AY =

u + Au

v.+..AV

Ay_u+Au_u _vAu - uAV

v+Av v v2+vAv

Au AV

AY_ vAx - uAx

Ax v2 +

vAVAx

Taking limits as Ax approaches zero gives

du do

dY dx - u dx

ax V2

and this proves (3.566).

The power formulas (3.563) and (3.564) remain to be proved, and we

deal with (3.563) first. Let y = x*. In case n = 0, we have y = 1 and

must prove that dy/dx = 0. This is true because if y = 1 for each x,

then Ay = 0, so Ay/Ax = 0 and hence dy/dx = 0. In case n = 1, we