Question
Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\sin\text{ xy}+\cos(\text{x}+\text{y})=1$
$\sin\text{ xy}+\cos(\text{x}+\text{y})=1$
✓
Answer
We have, $\sin\text{ xy}+\cos(\text{x}+\text{y})=1$Differentiating with respect to x,
$\frac{\text{d}}{\text{dx}}\sin\text{xy}+\frac{\text{d}}{\text{dx}}\cos(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\cos \text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})-\sin(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=0$
[Using chain rule]
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$
$\frac{\text{d}}{\text{dx}}\sin\text{xy}+\frac{\text{d}}{\text{dx}}\cos(\text{x}+\text{y})=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\cos \text{xy}\frac{\text{d}}{\text{dx}}(\text{xy})-\sin(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})=0$
[Using chain rule]
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]-\sin(\text{x}+\text{y})\Big[1+\frac{\text{dy}}{\text{dx}}\Big]=0$
$\Rightarrow\cos(\text{xy})\Big[\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}(1)\Big]-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}\cos(\text{xy})\frac{\text{dy}}{\text{dx}}+\text{y}\cos(\text{xy})-\sin(\text{x}+\text{y})+\sin(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\big[\text{x}\cos(\text{xy})+\sin(\text{x}+\text{y})\big]\frac{\text{dy}}{\text{dx}} \\ =\big[\sin(\text{x}+\text{y})-\text{y}\cos(\text{xy})\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big[\frac{\sin(\text{x}+\text{y})-\text{y}\cos\text{xy}}{\text{x}\cos\text{xy}+\sin(\text{x}+\text{y})}\Big]$
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