Page 103 - Foundations Of Differential Calculus

86 5. On the Differentiation of Algebraic Functions of One Variable
If we take a common denominator, we have

2
2
2
2
−x dx − a dx + x dx −a dx
√ = √ .
2
2
x 2 a − x 2 x 2 a − x 2
Hence the desired differential is
2
−a dx
dy = √ .
2
x 2 a − x 2
x 2
III. If y = √ ,welet
4
a + x 4
2 1
x = p and √ = q.
4
a + x 4
We find that
3
−2x dx
dp =2xdx and dq = 3/2 ,
4
4
(a + x )
so that
5 4
−2x dx 2xdx 2a xdx
pdq + qdp = 3/2 + √ = 3/2 .
4
4
4
4
4
(a + x ) a + x 4 (a + x )
It follows that the desired differential is
4
2a xdx
dy = √ .
4
4
4
(a + x ) a + x 4
x
IV. If y = √ ,welet
x + 1+ x 2
1
x = p and √ = q.
x + 1+ x 2
Since
dp = dx
and

√ √
−dx − (xdx) 1+ x 2 −dx x + 1+ x 2
dq = √ 2 √ 2 √
2 = 2 2
x + 1+ x x + 1+ x 1+ x
−dx
√ √ ,
= 2 2
x + 1+ x 1+ x