Question
Find $\frac{d y}{d x}$ if $ax + by^2 = \cos y$

Answer

It is given that $ax + by^2 = \cos y$
Differentiating both sides w.r.t. x, we get,
$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{by}^{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{cosy})$
$\Rightarrow \frac{d}{d x}(a x)+\frac{d}{d x}\left(b y^{2}\right)=\frac{d}{d x}(\cos y)$
$\Rightarrow a+b \frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\cos y)$
$\Rightarrow a+b \times 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x}$
$\Rightarrow(2 b y+\sin y) \frac{d y}{d x}=-a$
$\Rightarrow \frac{d y}{d x}=\frac{-a}{(2 b y+\sin y)}$

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