Question

Find $\frac {dy}{dx}$: $ax+by^2=cos\ y$

Answer 1

The given relationship is ax+by2 = cos y
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(ax+by2) = $\frac {d}{dx}$(cos y)

\implies$$\frac {d}{dx}(ax)+$\frac {d}{dx}$(by2) = $\frac {d}{dx}$(cosy)

$\implies$a+b$\frac {d}{dx}$(y2) = $\frac {d}{dx}$(cos y) …...…….(1)

Using chain rule,we obtain $\frac {d}{dx}$(y2) = 2y$\frac {dy}{dx}$ and $\frac {d}{dx}$(cosy) = -siny $\frac {dy}{dx}$ ………....(2)
From (1) and (2), we obtain
a+b.2y$\frac {dy}{dx}$ = -sin y.$\frac {dy}{dx}$

$\implies$(2by+sin y)$\frac {dy}{dx}$ = -a

∴ $\frac {dy}{dx}$ = $\frac {-a}{2by+sin\ y}$