If $\mathrm{y}=1+\frac{\alpha}{\left(\frac{1}{\mathrm{x}}-\alpha\right)}+\frac{\beta / \mathrm{x}}{\left(\frac{1}{\mathrm{x}}-\alpha\right)\left(\frac{1}{\mathrm{x}}-\beta\right)}+\frac{\gamma / \mathrm{x}^{2}}{\left(\frac{1}{\mathrm{x}}-\alpha\right)\left(\frac{1}{\mathrm{x}}-\beta\right)\left(\frac{1}{\mathrm{x}}-\gamma\right)}$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$
Given,
$\mathrm{y}=1+\frac{\alpha}{\left(\frac{1}{\mathrm{x}}-\alpha\right)}+\frac{\beta / \mathrm{x}}{\left(\frac{1}{\mathrm{x}}-\alpha\right)\left(\frac{1}{\mathrm{x}}-\beta\right)}+\frac{\gamma / \mathrm{x}^{2}}{\left(\frac{1}{\mathrm{x}}-\alpha\right)\left(\frac{1}{\mathrm{x}}-\beta\right)\left(\frac{1}{\mathrm{x}}-\gamma\right)}$
Using the theorem,
If $y=1+\frac{a x^{2}}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{c}{(x-c)}$, then, $\frac{d y}{d x}$
$=\frac{y}{x}\left\{\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right\}$
Here we have $\frac{1}{x}$ instead of $x$.
Hence, using the above theorem, we get,
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}\left[\frac{\alpha}{\frac{1}{\mathrm{x}}-\alpha}+\frac{\beta}{\frac{1}{\mathrm{x}}-\beta}+\frac{\gamma}{\frac{1}{\mathrm{x}}-\gamma}\right]$