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

Question

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

✓

Answer

Consider, $\sec(\text{x}+\text{y})=\text{xy}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \sec(\text{x}+\text{y})\tan(\text{x}+\text{y})+\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})\cdot\frac{\text{dy}}{\text{dx}}-\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\sec(\text{x}+\text{y})\tan(\text{x}+\text{y})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})-\text{x}\big]=\text{y}-\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})}{\sec(\text{x}+\text{y})\cdot\tan(\text{x}+\text{y})-\text{x}}$