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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
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Answer

Here,
$\text{xy }\log(\text{x}+\text{y})=1$
Differentiating with respect to x, we get
$\Rightarrow\frac{\text{d}}{\text{dx}}\big[\text{xy}\log(\text{x}+\text{y})\big]=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
[Using chain rule and product rule]
$\Rightarrow\text{xy}\times\Big(\frac{1}{\text{x}+\text{y}}\Big)\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Rightarrow\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
$\Big[\text{Since from equation (i)}\log(\text{x}+\text{y})=\frac{1}{\text{xy}}\Big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big)\Big(\frac{(\text{x}+\text{y})\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$
$=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$
So,
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}^2\text{y}+\text{x}+\text{y}}{\text{xy}^2+\text{x}+\text{y}}\Big)$

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