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If $y=1+\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+\ldots \infty$ with $|x|>1$, then $\frac{d y}{d x}=$
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The correct answer is:
$\frac{-y^{2}}{x^{2}}$
Given, $y=\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+\ldots \infty$
This is an infinite GP with $a=1$ and
$r=\frac{1}{x}$
$\because \quad S_{\infty}=\frac{a}{1-r}$
$y=\frac{1}{1-\frac{1}{x}}$
$y=\frac{x}{x-1}$
Now,
$\frac{d y}{d x}=\frac{(x-1)-x}{(x-1)^{2}}=\frac{x-1-x}{(x-1)^{2}}=\frac{-1}{(x-1)^{2}}$
$=-\frac{1}{\left(\frac{x^{2}}{y^{2}}\right)}=-\frac{y^{2}}{x^{2}}$
This is an infinite GP with $a=1$ and
$r=\frac{1}{x}$
$\because \quad S_{\infty}=\frac{a}{1-r}$
$y=\frac{1}{1-\frac{1}{x}}$
$y=\frac{x}{x-1}$
Now,
$\frac{d y}{d x}=\frac{(x-1)-x}{(x-1)^{2}}=\frac{x-1-x}{(x-1)^{2}}=\frac{-1}{(x-1)^{2}}$
$=-\frac{1}{\left(\frac{x^{2}}{y^{2}}\right)}=-\frac{y^{2}}{x^{2}}$
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