Page 131 - Foundations Of Differential Calculus

114 6. On the Differentiation of Transcendental Functions
IV. If
2x
y = arctan ,
1 − x 2
we let
2x
p = ,
1 − x 2
so that
2 2 2
2 1+ x 2dx 1+ x
1+ p = 2 and dp = 2 .
2
2
(1 − x ) (1 − x )
2
Hence, since dy = dp/ 1+ p , we have by the rule for tangents
(paragraph 197)
2dx
dy = 2 .
1+ x

V. If √
2
1+ x − 1
y = arctan ,
x
we let √
2
1+ x − 1
p = ,
x
so that
√
2
2 2+ x − 2 1+ x 2
p =
x 2
and
2
2 √ 2 2 √ 1+ x − 1 √ 1+ x 2
2+2x − 2 1+ x
2
1+ p = = .
x 2 x 2
But
√
2
−dx 2 dx dx 1+ x − 1
dp = √ x + 2 = √ .
x 2 1+ x 2 x x 2 1+ x 2
2
Hence, since dy = dp/ 1+ p ,wehave
dx
dy = .
2
2(1 + x )
From this we see that
√
2
1+ x − 1 1
arctan = arctan x.
x 2