Find dy dx when x and y are connected by the relation: (xy)+ x y …

Question

Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$

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Answer

Consider, $\sin(\text{xy})+\frac{\text{x}}{\text{y}}=\text{x}^2-\text{y}$$\Rightarrow\ \frac{\text{d}}{\text{dx}}(\sin\text{xy})+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)=\frac{\text{d}}{\text{dx}}\text{x}^2-\frac{\text{d}}{\text{dx}}\text{y}$
$\Rightarrow\ \cos\text{xy}\cdot\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{y}\frac{\text{d}}{\text{dx}}\text{x}-\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}}{\text{y}^2}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \cos\text{xy}\cdot\Big[\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\cdot\text{x}\Big]+\frac{\text{y}-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}=2\text{x}-\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{xy}+\frac{\text{y}}{\text{y}^2}\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \text{x}\cos\text{xy}\cdot\frac{\text{dy}}{\text{dx}}-\frac{\text{x}}{\text{y}^2}\frac{\text{dy}}{\text{dx}}+\frac{\text{dy}}{\text{dx}}=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\Big[\text{x}\cos\text{xy}-\frac{\text{x}}{\text{y}^2}+1\Big]=2\text{x}-\text{y}\cos\text{xy}-\frac{1}{\text{y}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\Big[\frac{2\text{xy}-\text{y}^2\cos\text{xy}-1}{\text{y}}\Big]\Big[\frac{\text{y}^2}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}\Big]$
$=\frac{2\text{xy}^2-\text{y}^3\cos\text{xy}-\text{y}}{\text{xy}^2\cos\text{xy}-\text{x}+\text{y}^2}$