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If y find If y=1+α(1x−α)+β/x(1x−α)(1x−β)+γ/x2(1x−α)(1x−β)(1x−γ), find dydx is: - Mathematics

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प्रश्न

\[\text{If y} = 1 + \frac{\alpha}{\left( \frac{1}{x} - \alpha \right)} + \frac{{\beta}/{x}}{\left( \frac{1}{x} - \alpha \right)\left( \frac{1}{x} - \beta \right)} + \frac{{\gamma}/{x^2}}{\left( \frac{1}{x} - \alpha \right)\left( \frac{1}{x} - \beta \right)\left( \frac{1}{x} - \gamma \right)}, \text{ find } \frac{dy}{dx}\] is:

विकल्प

  • `y (alpha/(alpha-x) + beta/(beta-x) + gamma/(gamma-x))`

  • `y/x (alpha/(1/(x-alpha)) + beta/(1/(x-beta)) + gamma/(1/(x-gamma)))`

  • `y (alpha/(1/(x-alpha)) + beta/(1/(x-beta)) + gamma/(1/(x-gamma)))`

  • `y/x ((alpha/x)/(1/(x-alpha)) + (beta/x)/(1/(x-beta)) + (gamma/x)/(1/(x-gamma)))`

MCQ
योग
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उत्तर

`y = (1/x)/(1/x - alpha) + (beta/x)/((1/x - alpha)(1/x - beta)) + (gamma/x^2)/((1/x - alpha)(1/x - beta)(1/x - gamma))`

`=(1/x^2)/((1/x - alpha)(1/x-beta)) + (gamma/x^2)/((1/x - alpha)(1/x-beta)(1/x - gamma))`

`y = (1/x^3)/((1/x - alpha)(1/x - beta))`

`=> log y = -3 log x (1/x - alpha) - log (1/x - beta) - log (1/x - gamma)`

`1/y dy/dx = (-3)/x - 1/(1/x - alpha) (-1/x - alpha) - 1/(1/x - beta) ((-1)/x^2) - 1/(1/x - gamma) (-1/x^2)`

`y = y/x [-3 + 1/x/1/x - alpha + 1/x/(1/x - beta) + 1/x/(1/x - gamma)]`

`y/x (alpha/(1/(x-alpha)) + beta/(1/(x-beta)) + gamma/(1/(x-gamma)))`

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अध्याय 11: Differentiation - Exercise 11.05 [पृष्ठ ९०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.05 | Q 61 | पृष्ठ ९०

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