Q. If $y = x + x^{2} + x^{3}\ldots\infty,\ where\ |x| < 1,then\ for\ |y| < 1,\frac{dx}{dy}$ is equal to
A) $y + y^{2} + y^{3} + ...\infty$
B) $1 - y + y^{2} - y^{3} + ...\infty$
C) $1 - 2y + 3y^{2} - ...\infty$
D) $1 + 2y + 3y^{2} + ...\infty$
Explanation
(c)
Given, $y = x + x^{2} + x^{3} + ... \Rightarrow y = \frac{x}{1 - x}$
$\Rightarrow \ x = \frac{y}{1 + y} = y - y^{2} + y^{3} - ...$
On differentiating w.r.t.$\ \ y,\$we get
$\frac{dx}{dy} = 1 - 2y + 3y^{2} - ...$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.