Page 135 - Foundations Of Differential Calculus

118 6. On the Differentiation of Transcendental Functions
since

tan x + tan dx
tan (x + dx)= ,
1 − tan x · tan dx
when the tangent is subtracted from this expression there remains
tan dx (1 + tan x · tan x)
dy = .
1 − tan x · tan dx

However, when the arc dx vanishes, the tangent is equal to the arc itself,
so that tan dx = dx, and the denominator 1 − dx tan x reduces to unity.
Hence

2
dy = dx 1 + tan x .
Since

2 2 1
1 + tan x = sec x = ,
2
cos x
we have
2 dx
dy = dx sec x = 2 .
cos x
We could also obtain this differential from the differentials of the sine and
cosine. Since tan x = sin x/cos x, we have (paragraph 164)
dx cos x · cos x + dx sin x · sin x dx
dy = =
2
2
cos x cos x
2 2
since sin x + cos x =1.
204. This differential can also be found in a different way. Since y = tan x,
we have x = arctan y, and by the rule given above,
dy
dx = .
1+ y 2
Since y = tan x,
1
2
1+ y = sec x = ,
cos x
2
so that dx = dy cos x and
dx
dy =
2
cos x
as before. The differential of the tangent of any arc is equal to the differ-
ential of the arc divided by the square of the cosine of the same arc.