if `y = x + x^(2) + x^(3) + …oo,` where |x| `lt 1,` then for `|y| lt 1, (dx)/(dy) ` is equal to

Options

A. `y + y^(2) + y^(3) + ….oo`
B. `1 - y + y^(2) - y^(3) + ….oo`
C. `1 - 2y + 3y^(2) - …oo`
D. `1 + 2y + 3y^(2) + ….oo`

Answer

C

Explanation

Given, `y = x + x^(2) + x^(3) + …oo`
`rArr y = (x)/(1-x)`
`rArr x = (y)/(1 + y) = y - y^(2) + y^(3) 0….oo`
On differentiating w.r.t. y, we get
` (dx)/(dy) = 1 -2 y + 3y^(2) 0…oo`