Continuity and Differentiability — Class 12 Maths Solution

We have, $xy = {e^{(x - y)}}$
By taking log on both sides , we get
$\log (xy) = \log {e^{(x - y)}}$

or log x + logy $= x - y$ ..(i)

Differentiating (i) on both sides w.r.t. x, we get

$\cfrac{1}{x} + \cfrac{1}{y}\cfrac{{dy}}{{dx}} = \left( {1 - \cfrac{{dy}}{{dx}}} \right) \Rightarrow \cfrac{{dy}}{{dx}}\left( {\cfrac{1}{y} + 1} \right) = 1 - \cfrac{1}{x}$

$\Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{{y(x - 1)}}{{x(y + 1)}}$