Page 129 - Foundations Of Differential Calculus

112 6. On the Differentiation of Transcendental Functions
and after substitution we obtain

dx
dy = .
1+ x 2
The differential of the arc whose tangent is given is equal to the differential
√
2
of the tangent divided by the square of the secant. We note that 1+ x is
the secant if x is the tangent.
198. In a similar way, suppose that the cotangent of an arc is given, so
that y is equal to the arccotangent of x. Since the tangent of that same arc
is 1/x,welet p =1/x, so that y = tan p. Since

dp −dx
dy = and dp = ,
1+ p 2 x 2
we make the substitutions to obtain
−dx
dy = ,
1+ x 2
that is, the differential of the arc of a cotangent is the negative of the
differential of the cotangent divided by the square of the cosecant.
If y is equal to the arcsecant of x, since

1
y = arccos ,
x
we have

dx dx
dy = = √ .
2
x 2 1 − 1/x 2 x x − 1
Also, if y is equal to the arccosecant of x, then
1
y = arcsin
x
and
−dx
dy = √ .
2
x x − 1
4
Frequently, the versed sine occurs. If y is equal to the versed sine of x,
√
2
since y = cos (1 − x), the sine of this arc is equal to 2x − x , so that
dx
dy = √ .
2x − x 2
4 The versed sine of α is equal to 1 − cos α.