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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
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Answer

Let $\text{y}=\text{x}^{\log\text{x}}+(\log\text{x}^\text{x})$
Also, let $\text{u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \text{y}=\text{v}+\text{u}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{dv}}{\text{dx}}+\frac{\text{du}}{\text{dx}}\ .....(\text{i})$
Now, $\text{u}=(\log\text{x})^\text{x}$
$\Rightarrow\log\text{u}=\log\big[(\log\text{x})^\text{x}\big]$
$\Rightarrow\log\text{u}=\text{x}\log(\log\text{x})$
Differentiating both sides with respect to x,
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\log(\log\text{x})\frac{\text{d}}{\text{dx}}(\text{x})+\text{x}\frac{\text{d}}{\text{dx}}\big[\log(\log\text{x})\big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\Big[\log(\log\text{x})+\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{\text{x}}{\log\text{x}}\times\frac{1}{\text{x}}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]\ .....(\text{ii})$
Also, $\text{v}=\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}^{\log\text{x}}$
$\Rightarrow\log\text{v}=\log\text{x}\log\text{x}=(\log\text{x})^2$
Differentiating both sides with respect to x,
$\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\big[(\log)^2\big]$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2(\log\text{x})\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{v}(\log\text{x})\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}\ .....(\text{iii})$
From (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=2\text{x}^{\log\text{x}}\frac{\log\text{x}}{\text{x}}+(\log\text{x})^\text{x}\Big[\log(\log\text{x})+\frac{1}{\log\text{x}}\Big]$

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