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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$
$\text{y}=\text{x}^{\sin\text{x}}+\big(\sin\text{x}\big)^\text{x}$
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Answer

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

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