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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate the functions given in Exercise:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
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Answer

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

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