Find $\frac{d y}{d x}$ when x and y are connected by the relation given.

$s i n (x y) + \frac{x}{y} = x^{2} - y$

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Solution

We have, $s i n (x y) + \frac{x}{y} = x^{2} - y$
On differentiating both sides. w.r.t. x, we get
$\frac{d}{d x} (s i n x y) + \frac{d}{d x} (\frac{x}{y}) = \frac{d}{d x} (x^{2}) - \frac{d y}{d x} \Rightarrow c o s (x y) . \frac{d}{d x} (x y) + \frac{y \frac{d}{d x} x - x . \frac{d}{d x} y}{y^{2}} = 2 x - \frac{d y}{d x} \Rightarrow c o s (x y) . [x . \frac{d y}{d x} + y] + \frac{y - x \frac{d y}{d x}}{y^{2}} = 2 x - \frac{d y}{d x} \Rightarrow x c o s (x y) . \frac{d y}{d x} + y c o s (x y) + \frac{y}{y^{2}} - \frac{x}{y^{2}} \frac{d y}{d x} = 2 x - \frac{d y}{d x} \Rightarrow \frac{d y}{d x} [x c o s (x y) - \frac{x}{y^{2}} + 1] = 2 x - y c o s (x y) - \frac{y}{y^{2}} ∴ \frac{d y}{d x} = [\frac{2 x y - y^{2} c o x y - 1}{y}] [\frac{y^{2}}{x y^{2} c o s x y - x + y^{2}}] = \frac{(2 x y - y^{2} c o s (x y) - 1)}{(x y^{2} c o s (x y) - x + y^{2})}$

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