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Question
If \[\sin \left( xy \right) + \frac{y}{x} = x^2 - y^2 , \text{ find} \frac{dy}{dx}\] ?
Solution
\[\text{ We have, }\sin\left( xy \right) + \frac{y}{x} = x^2 - y^2\]
Differentiating with respect to x, we get,
\[\Rightarrow \frac{d}{dx}\left( \sin xy \right) + \frac{d}{dx}\left( \frac{y}{x} \right) = \frac{d}{dx}\left( x^2 \right) - \frac{d}{dx}\left( y^2 \right)\]
\[ \Rightarrow \cos\left( xy \right)\frac{d}{dx}\left( xy \right) + \left\{ \frac{x\frac{dy}{dx} - y\frac{d}{dx}\left( x \right)}{x^2} \right\} = 2x - 2y\frac{dy}{dx} \]
\[ \Rightarrow \cos\left( xy \right)\left\{ x\frac{dy}{dx} + y\frac{d}{dx}\left( x \right) \right\} + \left\{ \frac{x\frac{dy}{dx} - y\left( 1 \right)}{x^2} \right\} = 2x - 2y\frac{dy}{dx}\]
\[ \Rightarrow \cos\left( xy \right)\left\{ x\frac{dy}{dx} + y\left( 1 \right) \right\} + \frac{1}{x^2}\left( x\frac{dy}{dx} - y \right) = 2x - 2y\frac{dy}{dx}\]
\[ \Rightarrow x \cos\left( xy \right)\frac{dy}{dx} + y \cos\left( xy \right) + \frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = 2x - 2y\frac{dy}{dx}\]
\[ \Rightarrow \frac{dy}{dx}\left\{ x \cos\left( xy \right) + \frac{1}{x} + 2y \right\} = \frac{y}{x^2} - y \cos\left( xy \right) + 2x\]
\[ \Rightarrow \frac{dy}{dx}\left\{ \frac{x^2 \cos\left( xy \right) + 1 + 2xy}{x} \right\} = \frac{1}{x^2}\left( y - x^2 y \cos\left( xy \right) + 2 x^3 \right)\]
\[ \Rightarrow \frac{dy}{dx} = \frac{2 x^3 + y - x^2 y \cos\left( xy \right)}{x\left( x^2 \cos\left( xy \right) + 1 + 2xy \right)}\]
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