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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
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Answer

Here,
$\text{xy}\log(\text{x}+\text{y})=1\ .....(\text{i})$
Differentiaitng with respect to x using chain rula, product rule,
$\frac{\text{dy}}{\text{dx}}(\text{xy}\log(\text{x}+\text{y}))=\frac{\text{d}}{\text{dx}}(1)$
$\text{xy}\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x})=0$
$\frac{\text{xy}}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{y}\log(\text{x}+\text{y})(1)=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\text{x}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}+\text{x}\log(\text{x}+\text{y})=0$
$\Big(\frac{\text{xy}}{\text{x}+\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\frac{\text{xy}}{\text{x}+\text{y}}+\text{x}\Big(\frac{1}{\text{xy}}\Big)\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{1}{\text{xy}}\Big)=0$
[Using equation (i)]
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}}{\text{x}+\text{y}}+\frac{1}{\text{y}}\Big]=-\Big[\frac{1}{\text{x}}+\frac{\text{xy}}{\text{x}+\text{y}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{xy}^2+\text{x}+\text{y}}{(\text{x}+\text{y})\text{y}}\Big]=-\Big[\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}(\text{x}+\text{y})}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big(\frac{\text{x}+\text{y}+\text{x}^2\text{y}}{\text{x}+\text{y}+\text{xy}^2}\Big)$

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