Question

Find $\frac {dy}{dx}$: $xy+y^2=tan\ x+y$

Answer 1

The given relationship is xy + y2 = tan x + y
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(xy+y2) = $\frac {d}{dx}$(tan x+y)

\implies$$\frac {d}{dx}(xy) + $\frac {d}{dx}$(y2) = $\frac {d}{dx}$(tan x) + $\frac {dy}{dx}$

$\implies$[y.$\frac {d}{dx}$(x) + x.$\frac {dy}{dx}$] + 2y$\frac {dy}{dx}$ = sec2x + $\frac {dy}{dx}$ [Using product rule and chain rule]

$\implies$y.1 + x.$\frac {dy}{dx}$ + 2y$\frac {dy}{dx}$ = sec2x + $\frac {dy}{dx}$

$\implies$(x + 2y -1)$\frac {dy}{dx}$= sec2x - y

∴ $\frac {dy}{dx}$ = $\frac {sec^2x - y}{(x+2y-1)}$