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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

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

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

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