Find $\frac{d y}{d x}$
$a x+b y^{2}=\cos y$
The given relationship is $a x+b y^{2}=\cos y$
Differentiating this relationship with respect to x, we obtain
$\frac{d}{d x}(a x)+\frac{d}{d x}\left(b y^{2}\right)=\frac{d}{d x}(\cos y)$
$\Rightarrow a+b \frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\cos y)$ ...(1)
Using chain rule, we obtain $\frac{d}{d x}\left(y^{2}\right)=2 y \frac{d y}{d x}$ and $\frac{d}{d x}(\cos y)=-\sin y \frac{d y}{d x}$ ...(2)
From (1) and (2), we obtain
$a+b \times 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x}$
$\Rightarrow(2 b y+\sin y) \frac{d y}{d x}=-a$
$\therefore \frac{d y}{d x}=\frac{-a}{2 b y+\sin y}$
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