Page 136 - Foundations Of Differential Calculus

6. On the Differentiation of Transcendental Functions 119
In a similar way if we let y = cot x, then x is equal to the arccotangent
of y and

−dy
dx = .
1+ y 2

But
1
2
1+ y = csc x = ,
sin x
2
so that dx = −dy sin x and
−dx
dy = 2 .
sin x

The differential of the cotangent of any arc is equal to the negative of the
differential of the arc divided by the square of the sine of the same arc.
Or since
cos x
cot x = ,
sin x
we have from the quotient rule

2 2
−dx sin x − dx cos x −dx
dy = 2 = 2
sin x sin x
as we have already seen.
205. If the secant of an arc is given, so that y = sec x, since y =1/cos x,
we have

dx sin x
dy = = dx tan x sec x.
2
cos x
In a similar way, if y = csc x,wehave

−dx cos x
dy = = −dx cot x csc x,
2
sin x
and it is not necessary to consider rules for special cases. If the versed sine
is given and y is equal to the versed sine of x, since y =1 − cos x,wehave
dy = dx sin x. In all of the cases in which some straight line is related to
a given arc, since it can always be expressed through a sine or a cosine, it
can always be differentiated without difficulty. This is true not only of the
first differentials, but also of the second and succeeding differentials by the
given rules. We let y = sin x, z = cos x, and we keep dx constant. Then we