For $|x|<1$, let $y=1+x+x^2+\ldots \infty$, then $\frac{d y}{d x}$ equals to
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For $|x|<1$, let $y=1+x+x^2+\ldots \infty$, then $\frac{d y}{d x}$ equals to

(A) $\frac{x}{y}$

(B) $\frac{x^2}{y^2}$

(C) $\frac{x}{y^2}$

(D) $x y^2$

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Best answer

The correct option will be (D).

HINT/SOLUTION —

$\because$ $y=1+x+x^2+\ldots \infty$

$\therefore \quad y=\frac{1}{1-x}=(1-x)^{-1}$

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