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: xx - 2sin x Let y = xx - 2sin x Let y = u - v $\Rightarrow$ u = xx and v = 2sin x For, u = xx Taking log on both sides, we get log u = log xx $\Rightarrow$ 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 = 2sin x Taking log on both sides, we get log v = log 2sin 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 $\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}$ $\frac{dy}{dx}$ = xx(1 + log x) - 2sin x.cosx.log 2
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