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Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation4 Marks
Question
Differentiate the following functions with respect to x:
$(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^\frac{1}{\text{x}}$
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Answer

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

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