Page 128 - Foundations Of Differential Calculus

6. On the Differentiation of Transcendental Functions 111
195. This differential of a circular arc can also more easily be found with-
out the aid of logarithms. If y = arcsin x, then x is the sine of the arc y,
that is, x = sin y. When we substitute x + dx for x, y becomes y + dy,so
that x + dx = sin (y + dy). Since
sin (a + b) = sin a · cos b + cos a · sin b,
we have
sin (y + dy) = sin y · cos dy + cos y · sin dy.

As dy vanishes the arc becomes equal to its sine, and its cosine becomes
equal to 1. For this reason sin (y + dy) = sin y + dy cos y, so that x + dx =
√
2
sin y + dy cos y. Since sin y = x,wehavecos y = 1 − x , and when these
√
2
values are substituted, we have dx = dy 1 − x , from which we obtain
dx
dy = √ .
1 − x 2
The arc of a given sine has a differential equal to the differential of the sine
divided by the cosine.
196. Suppose p is any function of x and that y is the arc whose sine is p,
2

that is, y = arcsin p. Since the differential of this arc is dy = dp/ 1 − p ,
2

where 1 − p expresses the cosine of that same arc, we can find the dif-
ferential of an arc whose cosine is given. If y = arccos x, then the sine
√ √
2
2
of this arc is equal to 1 − x , so that y = arcsin 1 − x . When we let
√ √
2
2
2
p = 1 − x , it follows that dp = −dx/ 1 − x and 1 − p = x, so that
−dx
dy = √ .
1 − x 2
The differential of the arc of a given cosine is equal to the negative of the
differential of the cosine divided by the sine of that same arc.
This result can also be shown in the following way. If y = arccos x,we
√
2
let z = arcsin x, so that dz = dx/ 1 − x . But the sum of the two arcs
y and z is equal to the constant 90 degrees, that is y + z is constant, so
that dy + dz =0, or dy = −dz. Hence we have the same result as before,
√
2
dy = −dx/ 1 − x .
197. If an arc whose tangent is given is to be differentiated, we begin
√
2
with y = arctan x. But then the sine is equal to x/ 1+ x and the cosine
√ √
2
2
is equal to 1/ 1+ x .Welet p = x/ 1+ x , so that
1
2

1 − p = √ ,
1+ x 2
2

and so y = arcsin p. Hence, by a rule already given dy = dp/ 1 − p . Since
√
2
p = x/ 1+ x ,wehave
dx
dp = 3/2 ,
2
(1 + x )