Continuity and Differentiability — Class 12 Maths Solution

Here,$x = 4t$ …(i)
$y = \cfrac{4}{t}$
Differentiating (i) \& (ii) w.r.t. t, we get

$\cfrac{{dx}}{{dt}} = 4$ and $\cfrac{{dy}}{{dt}} = \cfrac{{ - 4}}{{{t^2}}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{dy/dt}}{{dx/dt}} = \cfrac{{ - 4}}{{{t^2}}} \times \cfrac{1}{4} = \cfrac{{ - 1}}{{{t^2}}}$