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Differentiation — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferentiation5 Marks
Question
If $x^x + y^x = 1,$ prove that $\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}(\text{y}+\text{x}\log\text{y})}{\text{x}(\text{y}\log\text{x}+\text{x})}$
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Answer

Here,
$x^x + y^x = 1$
Taking on bith sides,
$\log(\text{x}^\text{y}\times\text{y}^\text{x})=\log(1)$
$\text{y}=\log\text{x}+\text{x}\log\text{y}=\log1$
$\big[\text{Since}, \log(\text{AB})=\log\text{A}+\log\text{B},\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
Differentiating it with respect to $x$ using product rule,
$\frac{\text{d}}{\text{dx}}(\text{y}\log\text{x})+\frac{\text{d}}{\text{dx}}(\text{x}\log\text{y})=\frac{\text{d}}{\text{dx}}(\log1)$
$\Big[\text{y}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\frac{\text{d}}{\text{dx}}(\log\text{y})+\log\text{y}\frac{\text{d}}{\text{dx}}\text{(x)}\Big]=0$
$\Big[\text{y}\Big(\frac{1}{\text{x}}\Big)\log\text{x}\frac{\text{dy}}{\text{dx}}\Big]+\Big[\text{x}\Big(\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}\Big)+\log\text{y}(1)\Big]=0$
$\frac{\text{y}}{\text{x}}+\log\text{x}\log\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}}\frac{\text{dy}}{\text{dx}}+\log\text{y}=0$
$\frac{\text{dy}}{\text{dx}}\Big(\log\text{x}+\frac{\text{x}}{\text{y}}\Big)=-\Big[\log\text{y}+\frac{\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}\Big[\frac{\text{y}\log\text{x}+\text{x}}{\text{y}}\Big]=-\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{x}}\Big]$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}\Big[\frac{\text{x}\log\text{y}+\text{y}}{\text{y}\log\text{x}+\text{x}}\Big]$

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