102] 
DERIVATIVES AND INTEGRALS 
205 
20. If 
ax 2 + 2 hxy + by 2 +2gx + 2fy + c=0, 
dy/dx= — {ax+hy+g)/{hx + by +/) 
d 2 y/dx 2 = — (ahc + 2fgh — af 2 — bg 2 — ch 2 )¡{lix -f by +f ) 3 . 
then 
and 
102. Some general theorems concerning derived func 
tions. In all that follows we suppose that <£ (x) is a function of 
x which has a derivative ft (x) for all values of x in question. 
This assumption of course involves the continuity of cp(x) (§ 92 (1)). 
The meaning of the sign of ft (x). Theorem A. If 
ft (x 0 ) > 0, (j){x) increases with x in the neighbourhood of x = x 0 : 
it is, moreover, an increasing function of x in the stricter sense 
(§ 88); that is to say we can find a small interval of values of x, 
stretching away from x 0 on both sides, and such that if x 1 and x 2 
are any values of x in this interval, and < x 2 , then <£(x x ) < </>(ic 2 ). 
For, as h-*-0, [(f) (x 0 + h) — <£ (x 0 )]/h converges to a positive limit. 
ft(x 0 ). This can only be the case if, for sufficiently small values of 
h, cj)(x 0 + h) — (¡)(x 0 ) and h have the same sign; and this is precisely 1 
what the theorem states. 
Of course from a geometrical point of view the result is 
intuitive, the inequality ft{x)>0 expressing the fact that the 
tangent to the curve y = (f>(x) makes a positive acute angle with 
the axis of x. 
Cor. 1. If ft(x) >0 for all values of x in a certain interval„ 
(f) (x) is an increasing f unction of x (in the stricter sense of § 88) 
throughout that interval. 
Cor. 2. If ft (x) > 0 throughout the interval {a, h), and 
f(a) = 0, then <f> (x) is positive throughout the interval (a, h). 
The reader should formulate for himself the corresponding 
theorems for the case in which ft(x) < 0. 
An immediate deduction from Theorem A is the following 
important theorem, generally known as Rolle’s Theorem : 
Theorem B. If <£ (a) = 0 and <£ (6) = 0 there must he at least 
one value of x which lies between a and b and for which ft (x) = 0. 
There are two possibilities: the first is that <£ (x) is equal to- 
zero throughout the whole interval (a, b). In this case ft(x) is 
also equal to zero throughout the interval. If this is not so there 
must be at least one value of x for which (f){x) is positive or 
negative, say positive. By Theorem 2 of § 89, there is a value