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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the functions given in Exercise:
$(\log\text{x})^\text{x}+\text{x}^{\log\text{x}}$
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Answer

Let $\text{y}=(\log\text{x})^\text{x}+\text{x}^{\log\text{x}}=\text{u}+\text{v }\text{ where u}=(\log\text{x})^\text{x}\text{ and v}=\text{x}^{\log\text{x}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}\ \dots\text{(i)}$
Now $\text{u}=(\log\text{x})^\text{x}\ \Rightarrow\ \log\text{u}=\log(\log\text{x})^\text{x}=\text{x}\log(\log\text{x})$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{u}=\frac{\text{d}}{\text{dx}}[\text{x}\log(\log\text{x})]\ \Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{\text{d}}{\text{dx}}[\log(\log\text{x})]+\log(\log\text{x})\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\log\text{x}}\frac{\text{d}}{\text{dx}}\log\text{x}+\log(\log\text{x}).1$
$\Rightarrow\ \frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\text{x}\frac{1}{\log\text{x}}\frac{1}{\text{x}}+\log(\log\text{x})\ \Rightarrow\ \frac{\text{du}}{\text{dx}}=\text{u}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]$
$\Rightarrow\ \frac{\text{du}}{\text{dx}}=(\log\text{x)}^\text{x}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]\ \dots\text{(ii)}$
Again $\text{v}=\text{x}^{\log\text{x}}\ \Rightarrow\ \log\text{v}=\log\text{x}^{\log\text{x}}=\log\text{x}.\log\text{x}=(\log\text{x})^2$
$\Rightarrow\ \frac{\text{d}}{\text{dx}}\log\text{v}=\frac{\text{d}}{\text{dx}}(\log\text{x})^2\ \Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2\log\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})$
$\Rightarrow\ \frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=2\log\text{x}.\frac{1}{\text{x}}\ \Rightarrow\ \frac{\text{dv}}{\text{dx}}=\text{v}\Big(\frac{2}{\text{x}}\log\text{x}\Big)=\text{x}^{\log\text{x}}\frac{2}{\text{x}}\log\text{x}$
$\Rightarrow\ \frac{\text{dv}}{\text{dx}}=2\text{x}^{\log\text{x}-1}\log\text{x}\ \dots\text{(iii)}$
Putting the values from eq. (ii) and (iii) in eq. (i),
$\frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1}{\log\text{x}}+\log(\log\text{x})\Big]+2\text{x}^{\log\text{x}-1}\log\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^\text{x}\Big[\frac{1+\log\text{x}\log(\log\text{x})}{\log\text{x}}\Big]+2\text{x}^{\log\text{x}-1}\log\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(\log\text{x})^{\text{x}-1}(1+\log\text{x}\log(\log\text{x}))+2\text{x}^{\log\text{x}-1}\log\text{x}$

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