Find dy dx in the following: xy + y^ 2 = + y - Vidyadip

Question

Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{xy} + \text{y}^{2} = \tan\text{x + y}$

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Answer

The given relationship is $\text{xy} + \text{y}^{2} = \tan\text{x + y}$ differenting this relationship with respect to x, we obtain $\frac{\text{d}}{\text{dx}}(\text{xy} + \text{y}^{2}) = \frac{\text{d}}{\text{dx}}(\tan\text{x} + \text{y})$ $\frac{\text{d}}{\text{dx}}(\text{xy})+\frac{\text{d}}{\text{dx}}\text{(y}^{2}) = \frac{\text{d}}{\text{dx}}(\tan\text{x})+\frac{\text{dy}}{\text{dx}}$ $\Rightarrow \Big[\text{y}.\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}.\frac{\text{dy}}{\text{dx}}\Big] +2\text{y} \frac{\text{dy}}{\text{dx}}=\sec^{2}\text{x}+\frac{\text{dy}}{\text{dx}}\ \ [\text{Using product rule and chain rule]}$$\Rightarrow(\text{x}+2\text{y}-1)\frac{\text{dy}}{\text{dx}}=\sec^{2}\text{x}-\text{y}$
$\therefore\frac{\text{dy}}{\text{dx}}= \frac{\sec^{2}\text{x}-\text{y}}{(\text{x} + 2\text{y}-1)}$

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