Page 120 - Foundations Of Differential Calculus

6. On the Differentiation of Transcendental Functions 103

1 1+ x
II. If y = ln , then
2 1 − x
1 1
y = ln (1 + x) − ln (1 − x) ,
2 2
so that
1 dx 1 dx dx
dy = 2 + 2 = .
1+ x 1 − x 1 − x 2
√
2
1 1+ x + x
III. If y = ln √ , since
2
2 1+ x − x
1 1
2
2
y = ln 1+ x + x − ln 1+ x − x ,
2 2
we have
1 1
2 dx 2 dx dx
dy = √ + √ = √ .
1+ x 2 1+ x 2 1+ x 2
This same result can be more easily obtained if we rationalize the denom-
√
2
inator by multiplying both numerator and denominator by 1+ x + x.
The result is
1 2
2
2
y = ln 1+ x + x =ln 1+ x + x ,
2
√
2
and as we have seen before, dy = dx/ 1+ x .
IV. If
√ √
1+ x + 1 − x
y =ln √ √ ,
1+ x − 1 − x
we let the numerator of this fraction be
√ √
1+ x + 1 − x = p
and the denominator
√ √
1+ x − 1 − x = q,
p

so that y =ln =ln p − ln q and dy = dp/p − dq/q. But
q
dx dx −dx √ √ −qdx
dp = √ − √ = √ 1+ x − 1 − x = √
2 1+ x 2 1 − x 2 1 − x 2 2 1 − x 2
and
dx dx pdx
dq = √ + √ = √ .
2 1+ x 2 1 − x 2 1 − x 2