Page 104 - Foundations Of Differential Calculus

5. On the Differentiation of Algebraic Functions of One Variable 87
we have
−xdx dx
√ √ + √
pdq + qdp = 2 2 2
x + 1+ x 1+ x x + 1+ x
√
2
dx 1+ x − x
√ √ .
= 2 2
x + 1+ x 1+ x
Therefore, the desired differential is
√
2
dx 1+ x − x
√ √ .
dy = 2 2
x + 1+ x 1+ x
If we multiply both numerator and denominator of this fraction by
√
2
1+ x − x,wehave
√
2 2 2
dx 1+2x − 2x 1+ x dx +2x dx
dy = √ = √ − 2x dx.
1+ x 2 1+ x 2
The same differential can be more easily obtained. Since
x
y = √ ,
x + 1+ x 2
√
2
if we multiply both numerator and denominator by 1+ x − x,we
have
2
2
4
2 2
y = x 1+ x − x = x + x − x .
By the previous rule we have
3
2
xdx +2x dx dx +2x dx
dy = √ − 2xdx = √ − 2x dx.
2
x + x 4 1+ x 2
V. If y =(a + x)(b − x)(x − c), then
dy =(a + x)(b − x) dx − (a + x)(x − c) dx +(b − x)(x − c) dx.

2
2
2 2 √
VI. If y = x a + x a − x , because of the three factors we have
2
2

x dx a + x 2
2
2
2
2
2 2 2
dy = dx a + x a − x +2x dx a − x − √
2
a − x 2
4 2 2 4
dx a + a x − 4x
= √ .
2
a − x 2