Question

Find $\frac{dy}{dx}$ of $ax+b{y}^2=\cos y$

Answer 1

$ax+b{y}^2=\cos y$
Differentiating both sides with respect to $x$, we obtain
$\frac{d}{dx}\left(ax\right)+\frac{d}{dx}\left(b{y}^2\right)=\frac{d}{dx}\left(\cos y\right)$
$\Rightarrow a+b\times 2y\frac{dy}{dx}=-\sin y\frac{dy}{dx}$
$\Rightarrow (2by+\sin y)\frac{dy}{dx}=-a$
$\therefore \frac{dy}{dx}=\frac{-a}{2by+\sin y}$