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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