254 Proceedings of the Royal Society of Edinburgh. [Sess. 
This formula gives the explicit expression of the differential coefficient 
of the continued-fraction 
1 
b + x 
Zq -f- x ■ 
b n + x 
where a 4 = c 1 d 1 , a 2 = c 2 d 2 , . . . a n = c n d n . 
It will be noticed that the determinant in the numerator is the principal 
minor of the determinant in the denominator: and also that the latter 
determinant is the square of the determinant formed of the elements in its 
top right-hand quadrant. We may further remark that by reducing the 
order of the determinants in the same way as in a well-known proof of 
the multiplication-theorem for determinants, we can obtain the result in 
the form 
(x + bf) 2 + c x d x + c 2 cZ 2 
df2x + Zq -p Z> 2 ) 
d%d 3 
0 
6 * 2 ( 2 ^ + Zq + Zq) 
(x -P Zq) 2 -P c 2 d 2 + c 3 d 3 
dffiX + b 2 + b s ) 
d 3 d 4 
C 2 C 3 
c 3 ( 2x -p Zq + Zq) 
(x + b 3 ) 2 -p c 3 d 3 + c 4 eZ 4 
cZ 4 ( 2x -P Zq -P Zq) 
0 
c 3 c 4 . . . 
c 4 (2a? + Zq + Zq) . . . 
{x + Zq) 2 + c 4 d 4 -P c h d b . . . 
(x + b) 2 + c Y d x 
\ 
\(2x + b + bf) 
c l c 2 
0 
dffXx + b + Zq) 
(x -P Zq) 2 -P c 4 cZ 4 -P c 2 cZ 2 
cff2x -p Zq -p Zq) 
C 2 Cg . . . 
dfl 2 
dfflx + Zq + Zq) 
(x -P b 2 ) 2 + c 2 d 2 + c 3 d 3 
c 3 (2x + Zq + Zq) . . . 
0 
d^d 3 
d 3 (2x -p Zq + Zq) 
(x -P b 3 ) 2 + c 3 d 3 + c 4 cZ 4 . . . 
For the benefit of those who dislike or distrust symbolic- matrix proofs, the following 
proof by ordinary “mathematical induction” may be added. 
First, the theorem may be verified by direct expansion in the simplest cases. Suppose 
then that formula (19) holds when there is some particular number n of quotients in the 
continued-fraction, say for the continued-fraction 
T 
1 
7 ~ 
h ' + x ~bfVx- 
On , 
bn + X 
then we shall prove that it holds for the next higher number (?i + l) of quotients, say for 
the continued-fraction 
