Continuity and Differentiability — Class 12 Maths Solution

and $y = t - \frac{1}{t}$
therefore,$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {t + \frac{1}{t}} \right)$ and $\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {t - \frac{1}{t}} \right)$

$\Rightarrow$ $\frac{{dx}}{{dt}} = 1 + ( - 1){t^{ - 2}}$ and $\frac{{dy}}{{dt}} = 1 - ( - 1){t^{ - 2}}$

$\Rightarrow$ $\frac{{dx}}{{dt}} = 1 - \frac{1}{{{t^2}}}$ and $\frac{{dy}}{{dt}} = 1 + \frac{1}{{{t^2}}}$

$\Rightarrow$ $\frac{{dx}}{{dt}} = \frac{{{t^2} - 1}}{{{t^2}}}$ and $\frac{{dy}}{{dt}} = \frac{{{t^2} + 1}}{{{t^2}}}$

therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{{t^2} + 1/{t^2}}}{{{t^2} - 1/{t^2}}} = \frac{{{t^2} + 1}}{{{t^2} - 1}}$

$\frac{dy}{dx} = \frac{{{t^2} + 1}}{{{t^2} - 1}}$