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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the function given in Exercise:
$(\text{x}\cos\text{x})^\text{x}+(\text{x}\sin\text{x})^{\frac{1}{\text{x}}}$
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Answer

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

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