Continuity and Differentiability — Maths STD 12 Science — Question

$\therefore \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4\frac{d}{{dt}}t = 4$

and $\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{t}} \right) = \frac{{t\frac{d}{{dt}}4 - 4\frac{d}{{dt}}t}}{{{t^2}}}$

$\Rightarrow \frac{{dy}}{{dt}} = \frac{{t \times 0 - 4 \times 1}}{{{t^2}}} = - \frac{4}{{{t^2}}}$

Now $\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - \frac{4}{{{t^2}}}}}{4} = \frac{{ - 1}}{{{t^2}}}$