If y=1+α(1x−α)+β/x(1x−α)(1x−β)+γ/x2(1x−α)(1x−β)(1x−γ) , then dydx is

y=1x1x−α+βx1x−α1x−β+xγ21x−α1x−β1x−γ=x121x−α1x−β+xy21x−α1x−β1x−α=x131x−α1x−β1x−γ log⁡y=−3ln⁡x−ln⁡1x−α−ln⁡1x−β−ln⁡1x−γ1yy1=−3x+x121x−α+x121x−β+x121x−γy1=yx−3+1x1x−α+1x1x−β+1x1x−γy1=yxα1/x−α+β1/x−β+γ1/x−y

If y=1+α(1x−α)+β/x(1x−α)(1x−β)+γ/x2(1x−α)(1x−β)(1x−γ) , then dydx is

y=1x1x−α+βx1x−α1x−β+xγ21x−α1x−β1x−γ=x121x−α1x−β+xy21x−α1x−β1x−α=x131x−α1x−β1x−γ log⁡y=−3ln⁡x−ln⁡1x−α−ln⁡1x−β−ln⁡1x−γ1yy1=−3x+x121x−α+x121x−β+x121x−γy1=yx−3+1x1x−α+1x1x−β+1x1x−γy1=yxα1/x−α+β1/x−β+γ1/x−y

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If $y = 1 + \frac{α}{(\frac{1}{x} - α)} + \frac{β / x}{(\frac{1}{x} - α) (\frac{1}{x} - β)} + \frac{γ / x^{2}}{(\frac{1}{x} - α) (\frac{1}{x} - β) (\frac{1}{x} - γ)}$ , then $\frac{d y}{d x}$ is

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$y (\frac{α}{α - x} + \frac{β}{β - x} + \frac{γ}{γ - x})$

$\frac{y}{x} (\frac{α}{1 / x - α} + \frac{β}{1 / x - β} + \frac{γ}{1 / x - γ})$

$y (\frac{α}{1 / x - α} + \frac{β}{1 / x - β} + \frac{γ}{1 / x - γ})$

$\frac{y}{x} (\frac{α / x}{1 / x - α} + \frac{β / x}{1 / x - β} + \frac{γ / x}{1 / x - γ})$

answer is B.

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Detailed Solution

$y = \frac{\frac{1}{x}}{\frac{1}{x} - α} + \frac{\frac{β}{x}}{(\frac{1}{x} - α) (\frac{1}{x} - β)} + \frac{x^{\frac{γ}{2}}}{(\frac{1}{x} - α) (\frac{1}{x} - β) (\frac{1}{x} - γ)}$

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

$\log y = - 3 \ln x - \ln (\frac{1}{x} - α) - \ln (\frac{1}{x} - β) - \ln (\frac{1}{x} - γ)$

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