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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
If $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^\text{x}}}\times\text{x}^{\text{e}^{\text{x}}}\Big\{\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{e}^{\text{e}^\text{x}}\Big\{\frac{1}{\text{x}}+\text{e}^\text{x}\times\log\text{x}\Big\}+\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^{\text{e}}}\times\text{x}^{\text{e}-1}\Big\{\text{x}+\text{e}\log\text{x}\Big\}$
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Answer

We have, $\text{y}=\text{e}^{\text{x}^{\text{e}^\text{x}}}+\text{x}^{\text{e}^{\text{e}^\text{x}}}+\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\text{y}=\text{u}+\text{v}+\text{w}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}+\frac{\text{dv}}{\text{dx}}+\frac{\text{dw}}{\text{dx}}\ .....(\text{i})$
Where $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}},\text{v}=\text{x}^{\text{e}^{\text{e}^{\text{x}}}}\text{ and w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
Now, $\text{u}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\ .....(\text{ii})$
Taking log on both sides,
$\log\text{u}=\log\text{e}^{\text{x}^{\text{e}^{\text{x}}}}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\log\text{e}$
$\Rightarrow\log\text{u}=\text{x}^{\text{e}^\text{x}}\ .....(\text{iii})$
Taking $\log$ on both sides,
$\log\log\text{u}=\log\text{x}^{\text{e}^\text{x}}$
$\Rightarrow\log\log\text{u}=\text{e}^\text{x}\log\text{x}$
Differentiating with respect to x,
$\Rightarrow\frac{1}{\log\text{u}}\frac{\text{d}}{\text{dx}}(\log\text{u})=\text{e}^\text{x}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{1}{\log\text{u}}\frac{1}{\text{u}}\frac{\text{du}}{\text{dx}}=\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{u}\log\text{u}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]$
$\Rightarrow\frac{\text{du}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{A})$
[Using equation (ii) and (iii)]
Now, $\text{v}=\text{x}^{\text{e}^{\text{e}^\text{x}}}\ .....(\text{iv})$
Taking log on both sides,
$\log\text{v}=\log\text{x}^{\text{e}^{\text{e}^\text{x}}}$
$\Rightarrow\log\text{v}=\text{e}^{\text{e}^\text{x}}\log\text{x}$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}\big(\text{e}^{\text{e}^\text{x}}\big)$
$\Rightarrow\frac{1}{\text{v}}\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\frac{\text{d}}{\text{dx}}(\text{e}^\text{x})$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{v}\Big[\text{e}^{\text{e}^\text{x}}\big(\frac{1}{\text{x}}\big)+\log\text{xe}^{\text{e}^\text{x}}\text{e}^\text{x}\Big]$
$\Rightarrow\frac{\text{dv}}{\text{dx}}=\text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big]\ .....(\text{B})$
[Using equation (4)]
Now, $\text{w}=\text{e}^{\text{x}^{\text{x}^{\text{e}}}}\ .....(\text{v})$
Taking log on sides,
$\log\text{w}=\log\text{e}^{\text{x}^{\text{x}^{\text{e}}}}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^\text{e}}\log\text{e}$
$\Rightarrow\log\text{w}=\text{x}^{\text{x}^{\text{e}}}\ .....(\text{vi})$
Taking log on both sides,
$\log\log\text{w}=\log\text{x}^{\text{x}^{\text{e}}}$
$\Rightarrow\log\log\text{w}=\text{x}^{\text{e}}\log\text{x}$
$\Rightarrow\frac{1}{\log\text{w}}\frac{\text{d}}{\text{dx}} (\log\text{w})=\text{x}^\text{e}\frac{\text{d}}{\text{dx}}(\log\text{x})+\log\text{x}\frac{\text{d}}{\text{dx}}(\text{x}^\text{e})$
$\Rightarrow\frac{1}{\log\text{w}}\big(\frac{1}{\text{w}}\big)\frac{\text{dw}}{\text{dx}}=\text{x}^{\text{e}}\big(\frac{1}{\text{x}}\big)\log\text{xex}^{\text{e}-1}$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{w}\log\text{w}\big[\text{x}^{\text{e}-1}+\text{e}\log\text{xx}^{\text{e}-1}\big]$
$\Rightarrow\frac{\text{dw}}{\text{dx}}=\text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})\ .....(\text{C})$
Using equation (A), (B) and (C) in equation (i), we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^{\text{e}^{\text{x}}}}\times\text{x}^{\text{e}^\text{x}}\Big[\frac{\text{e}^\text{x}}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{e}^{\text{e}^\text{x}}}\times\text{e}^{\text{e}^\text{x}}\Big[\frac{1}{\text{x}}+\text{e}^\text{x}\log\text{x}\Big] + \text{e}^{\text{x}^{\text{x}^\text{e}}}\text{x}^{\text{x}^\text{e}}\text{x}^{\text{e}-1}(1+\text{e}\log\text{x})$

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