From this it appears, that having calculated the logarithms of numbers to any one base, we may easily find the logarithms of the same numbers to any other base.

From the preceding propositions, we may find the means of differentiating functions involving variable indices.

To find the differential of y, where (y) and (*) represent some functions of (x).

.. dw=y (hyp. log. y.dz + zdy).

COR. 1. If y=x=x, we have

d.xxx dx {hyp. log. (x)+1}.

COR. 1. If the value of (x) be such, that 1

du

hyp. log. (x)=0, it is manifest, that = 0; and,

dx

therefore, u is a maximum; but, in this case,

hyp. log. x=1= hyp, log. e ; .'. x=€.

To find the differential of hyp. log. (1+x), and

=

d. hyp. log. (a+x)-d. hyp. log. (a – x)

= d. hyp. log. (x-a) - d. hyp. log. (x+a)

= d. h. l. (a— √√ a2 — x2) − d. h.l. (a + √ a2 − x2)

=

xdx

-

+

xdx

'a2 — x2. (a — √√ a2 — x2) √ a2x2 (a + √ a2 — x2)

2 adx

x √ a2-x2

=d.h.l. (√√ a2 + x2 — a) — d. h. l. (√ a2+x2 + a)

=

=

xdx

xdx

a2+x2+a)

√ a2 + x2 (√ a2+x2-α) √ a2+x3 (√ a2 + x2 + a)

2 adx

x √ a2+x2

These six preceding forms will, hereafter, be useful in integration.

The differentiation of algebraic functions is sometimes rendered easier, by first taking their hyperbolic logarithms, and then differentiating ; thus, if

From the preceding propositions, we may also find the values of vanishing fractions, whose numerators and denominators involve variable exponentials.

The differential coefficient of the numerator= Aa, and when x=0, it= A. Also the differential coefficient of the denominator=1, whatever be the

hyp. log. (a).

must = A =

x

Ex. 2. To find the value of hyp. log. (1+x)

when a = 0.

=

The differential coefficient of hyp. log. (1+x)

1

1 + x

,

and when x=0, it becomes unity. Also

the differential coefficient of (x) always equals

unity; hence, the value of

hyp. log. (1+x)

when

x=0, is unity.

Σ