Question

$\mathrm{If}y=1+\frac{\alpha }{\left({\displaystyle \frac{1}{x}-\alpha }\right)}+\frac{{\displaystyle \raisebox{1ex}{$\beta $}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}{\left({\displaystyle \frac{1}{x}-\alpha }\right)\left({\displaystyle \frac{1}{x}-\beta }\right)}+\frac{{\displaystyle \raisebox{1ex}{$\gamma $}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}}{\left({\displaystyle \frac{1}{x}-\alpha }\right)\left({\displaystyle \frac{1}{x}-\beta }\right)\left({\displaystyle \frac{1}{x}-\gamma }\right)},\mathrm{find}\frac{dy}{dx}.$

Answer 1

$$\begin{gathered} y=1+\frac{\alpha }{\left({\displaystyle \frac{1}{x}}-\alpha \right)}+\frac{{\displaystyle \raisebox{1ex}{$\beta $}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}{\left({\displaystyle \frac{1}{x}}-\alpha \right)\left({\displaystyle \frac{1}{x}}-\beta \right)}+\frac{{\displaystyle \raisebox{1ex}{$\gamma $}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}}{\left({\displaystyle \frac{1}{x}}-\alpha \right)\left({\displaystyle \frac{1}{x}}-\beta \right)\left({\displaystyle \frac{1}{x}}-\gamma \right)} \\ \Rightarrow y=1+\frac{\alpha x}{1-\alpha x}+\frac{\beta x}{\left(1-\alpha x\right)\left(1-\beta x\right)}+\frac{\gamma x}{\left(1-\alpha x\right)\left(1-\beta x\right)\left(1-\gamma \right)} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log y=\log 1+\log \alpha x-\log \left(1-\alpha x\right)+\log \beta x-\log \left(1-\alpha x\right)-\log \left(1-\beta x\right)+\log \gamma x-\log \left(1-\alpha x\right)-\log \left(1-\beta x\right)-\log \left(1-\gamma x\right) \\ \Rightarrow \log y=\log \alpha x-3\log \left(1-\alpha x\right)+\log \beta x-2\log \left(1-\beta x\right)+\log \gamma x-\log \left(1-\gamma x\right) \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{3\alpha }{1-\alpha x}+\frac{1}{x}+\frac{2\beta }{1-\beta x}+\frac{1}{x}+\frac{\gamma }{1-\gamma x} \\ \Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\left[3+\frac{3\alpha }{\left({\displaystyle \frac{1}{x}-\alpha }\right)}+\frac{2\beta }{\left({\displaystyle \frac{1}{x}-\beta }\right)}+\frac{\gamma }{\left({\displaystyle \frac{1}{x}-\gamma }\right)}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left[\frac{\alpha }{\left({\displaystyle \frac{1}{x}-\alpha }\right)}+\frac{\beta }{\left({\displaystyle \frac{1}{x}-\beta }\right)}+\frac{\gamma }{\left({\displaystyle \frac{1}{x}-\gamma }\right)}\right] \end{gathered}$$