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Differentiation — Maths STD 12 Science — Question

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

Here,
$x^x + y^x = 1$
$\text{e}^{\log\text{x}^\text{x}}+\text{e}^{\log\text{y}^\text{x}}=1$
$\text{e}^{\text{x}\log\text{x}}+\text{e}^{\text{x}\log\text{y}}=1$
$\big[\text{Since},\text{e}^{\log\text{a}}=\text{a}.\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to x using product rule and chain rule,
$\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{x}}\big)+\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{x}\log\text{y}}\big)=\frac{\text{d}}{\text{dx}}(1)$
$\text{e}^{\text{x}\log\text{x}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{x})+\text{e}^{\text{x}\log\text{y}}\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=0$
$\text{e}^{\text{x}\log\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x})\Big] \\ +\text{e}^{\log\text{y}^\text{x}}\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}(\text{x})\Big]=0$
$\text{x}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{x}}\Big)+\log\text{x}(1)\Big]+\text{y}^\text{x}\Big[\text{x}\Big(\frac{1}{\text{y}}\Big)\frac{\text{dy}}{\text{dx}}+\log\text{y}(1)\Big]=0$
$\text{x}^\text{x}[1+\log\text{x}]+\text{y}^\text{x}\Big(\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}\Big)=0$
$\text{y}^\text{x}\times\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\big(\text{xy}^{\text{x}-1}\big)\frac{\text{dy}}{\text{dx}}=-\big[\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}\big]$
$\frac{\text{dy}}{\text{dx}}=-\Big[\frac{\text{x}^\text{x}(1+\log\text{x})+\text{y}^\text{x}\log\text{y}}{\text{xy}^{\text{x}-1}}\Big]$

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