Question

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

Answer 1

`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)))`