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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
Differentiate the following functions with respect to x:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
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Answer

Let $\text{y}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}+\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
Also, let $\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$ and $\text{v}=\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
$\therefore\text{y}=\text{u}+\text{v}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}....(1)$
Then, $\text{u}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$
$\Rightarrow\log\text{u}=\log\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}$
$\Rightarrow\log\text{u}=\text{x}\log\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Differentiating both sides with respect to x, we obtain,
$\frac{1}{\text{u}}\cdot\frac{\text{du}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\text{x})\times\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\text{x}\times\frac{\text{d}}{\text{dx}}\Big[\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]$
$\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=1\times\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\text{x}\times\frac{1}{\Big(\text{x}+\frac{1}{\text{x}}\Big)}\cdot\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\log\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{\text{x}^2-1}{\text{x}^2+1}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Big[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big]$
$\text{v}=\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}$
$\Rightarrow\log\text{v}=\log\Bigg[\text{x}^{\Big(1+\frac{1}{\text{x}}\Big)}\Bigg]$
$\Rightarrow\log\text{v}=\Big(1+\frac{1}{\text{x}}\Big)\log\text{x}....(2)$
Differentiating both sides with respect to x, we obtain
$\frac{1}{\text{v}}\cdot\frac{\text{dv}}{\text{dx}}=\Big[\frac{\text{d}}{\text{dx}}\Big(1+\frac{1}{\text{x}}\Big)\Big]\times\log\text{x}+\Big(1+\frac{1}{\text{x}}\Big)\cdot\frac{\text{d}}{\text{dx}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\Big(-\frac{1}{\text{x}^2}\Big)\log\text{x}+\Big(1+\frac{1}{\text{x}}\Big)\cdot\frac{1}{\text{x}}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=-\frac{\log\text{x}}{\text{x}^2}+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\frac{-\log\text{x}+\text{x}+1}{\text{x}^2}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{x}^{\big(1+\frac{1}{\text{x}}\big)}\Big(\frac{\text{x}+1-\log\text{x}}{\text{x}^2}\Big) .....(3)$
Therefore, from (i), (ii) and (iii), we obtain
$\frac{\text{dy}}{\text{dx}}=\Big(\text{x}+\frac{1}{\text{x}}\Big)^\text{x}\Bigg[\frac{\text{x}^2-1}{\text{x}^2+1}+\log\Big(\text{x}+\frac{1}{\text{x}}\Big)\Bigg]+\text{x}^{\big(1+\frac{1}{\text{x}}\big)}\Big(\frac{\text{x+1}-\log\text{x}}{\text{x}^2}\Big)$

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