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If $y = 1 + \frac{c_{1}}{x - c_{1}} + \frac{c_{2} x}{(x - c_{1}) (x - c_{2})} + \frac{c_{3} x^{2}}{(x - c_{1}) (x - c_{2}) (x - c_{3})}$ then $\frac{d y}{d x}$ is equal to

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$\frac{- y}{x} [\frac{c_{1}}{c_{1} - x} + \frac{c_{2}}{x} + \frac{c_{3}}{c_{3} - x}]$

$\frac{- y}{x} [\frac{c_{1}}{x} + \frac{c_{2}}{x} + \frac{c_{3}}{c_{3} - x}]$

$\frac{y}{x} [\frac{c_{1}}{c_{1} - x} + \frac{c_{2}}{c_{2} - x} + (\frac{c_{3}}{- x})]$

$\frac{y}{x} [\frac{c_{1}}{c_{1} - x} + \frac{c_{2}}{c_{2} - x} + \frac{c_{3}}{c_{3} - x}]$

answer is D.

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Detailed Solution

$y = \frac{x}{x - c_{1}} + \frac{c_{2} x}{(x - c_{1}) (x - c_{2})} + \frac{c_{3} x^{2}}{(x - c_{1}) (x - c_{2}) (x - c_{3})}$

$= \frac{x^{2}}{(x - 4 c_{1}) (x - c_{2})} + \frac{c_{3} x^{2}}{(x - c_{1}) (x - c_{2}) (x - c_{3})}$

$= \frac{x^{3}}{(x - c_{1}) (x - c_{2}) (x - c_{3})}$

$∴ \ln y = \ln x^{3} - \ln (x - c_{1}) - \ln (x - c_{2}) - \ln (x - c_{3})$

$∴ \ln y = 3 \ln x - \ln (x - c_{1}) - \ln (x - c_{2}) - \ln (x - c_{3})$

$\frac{y^{'}}{y} = \frac{3}{x} - \frac{1}{x - c_{1}} - \frac{1}{x - c_{2}} - \frac{1}{x - c_{3}}$

$y^{'} = \frac{y}{x} [3 - \frac{x}{x - c_{1}} - \frac{x}{x - c_{2}} - \frac{x}{x - c_{3}}] = \frac{y}{x} [\frac{c_{1}}{c_{1} - x} + \frac{c_{2}}{c_{2} - x} + \frac{c_{3}}{c_{3} - x}]$

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