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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $x^y + y^x = (x + y)^{x+y}$, find $\frac{\text{dy}}{\text{dx}}$
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Answer

Here,
$x^y + y^x = (x + y)^{x+y}$
$\text{e}^{\log\text{x}^\text{y}}+\text{e}^{\log\text{y}^\text{x}}=\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}$
$\text{e}^{\text{y}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=\text{e}^{{(\text{x}+\text{y})}\log(\text{x}+\text{y})}$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using chain rule, product rule,
$\Rightarrow\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{y}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big] \\ +\text{e}^{\text{x}\log\text{y}}\Big[\text{x}\frac{\text{d}}{\text{dx}}\log\text{y}+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ =\text{e}^{(\text{x}+\text{y})\log(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}\big[(\text{x}+\text{y})\log(\text{x}+\text{y})\big]$
$\Rightarrow\text{e}^{\text{y}\log\text{x}}\Big[\text{y}\big(\frac{1}{\text{x}}\big)+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{e}^{\log\text{x}}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big] \\ =\text{e}^{\log(\text{x}+\text{y})^{(\text{x}+\text{y})}}\Big[(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}\log(\text{x}+\text{y})+\log(\text{x}+\text{y})\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})\Big]$
$\Rightarrow\text{x}^\text{y}\Big[\frac{\text{y}}{\text{x}}+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\text{y}^\text{x}\Big[\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big] \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[(\text{x}+\text{y})\frac{1}{(\text{x}+\text{y})}\frac{\text{d}}{\text{dx}}(\text{x}+\text{y})+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^\text{y}\times\frac{\text{y}}{\text{x}}+\text{x}^{\text{y}}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\text{y}^\text{x}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}\Big[1\times\Big(1+\frac{\text{dy}}{\text{dx}}\Big)+\log(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)\Big]$
$\Rightarrow\text{x}^{\text{y}-1}\times\text{y}+\text{x}^\text{y}\log\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}-1}\times\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}^{\text{x}}\log\text{y} \\ =(\text{x}+\text{y})^{(\text{x}+\text{y})}+(\text{x}+\text{y})^{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}} \\ +(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})+(\text{x}+\text{y})^{(\text{x}+\text{y})}\log(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\Big[\text{x}^{\text{y}}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))\Big] \\ =(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^{\text{x}}\log\text{y}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\bigg[\frac{(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))-\text{x}^{\text{y}-1}\times\text{y}-\text{y}^\text{x}\log\text{y}}{\text{x}^\text{y}\log\text{x}+\text{xy}^{\text{x}-1}-(\text{x}+\text{y})^{\text{x}+\text{y}}(1+\log(\text{x}+\text{y}))}\bigg]$

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