Question

$Find$ $\frac {dy}{dx}$: $2x+3y=sin \ y$

Answer 1

The given relationship is 2x + 3y = sin y
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(2x + 3y) = $\frac {d}{dx}$(sin y)
\implies$$\frac {d}{dx}(2x) + $\frac {d}{dx}$(3y) = $\frac {d}{dx}$(sin y)
$\implies$2 + 3$\frac {dy}{dx}$ = cos y. $\frac {dy}{dx}$
$\implies$2 = (cos y - 3).$\frac {dy}{dx}$
∴$\frac {dy}{dx}$ = $\frac {2}{cos\ y-3}$