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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
Differentiate the function $x^{x}-2^{\sin x}$ w.r.t. x.
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Answer

Given: $x^x - 2 \sin x$
Let $y = x^x - 2 \sin x$
Let $y = u - v$
$\Rightarrow u = x^x$ and $v = 2 \sin x$
For, $u = x^x$
Taking \log on both sides, we get
$\log u = \log x^x$
$\Rightarrow \log u = x. \log(x)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\mathrm{x})]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{u}\left[\mathrm{x} \cdot \frac{1}{\mathrm{x}}+\log \mathrm{x} \cdot(1)\right]$
$\Rightarrow \frac{d u}{d x}=x^{x}(1+\log x)$
For, $v = 2^{\sin x}$
Taking \log on both sides, we get
$\log v = \log 2^{\sin x}$
$\Rightarrow \log v = \sin x. \log (2)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{v})=\frac{\mathrm{d}}{\mathrm{dx}}[\sin \mathrm{x} \cdot \log (2)]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\log 2 \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})$
$\Rightarrow \frac{d v}{d x}=v[\log 2 .(\cos x)]$
$\Rightarrow \frac{d v}{d x}=2^{\sin x} \cdot \cos x \cdot \log 2$
Because,$y = u - v$
$\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$
$\frac{dy}{dx} = x^x(1 + \log x) - 2^{\sin x }\cos x \log 2$

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