Differentiate the function y = x^{x} - 2^{sin | vedclass

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Let $y = x^{x} - 2^{\sin x}$.Let $u = x^{x}$ and $v = 2^{\sin x}$.Then $y = u - v$,so $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.For $u = x^{x}$,

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Let $y = x^{x} - 2^{\sin x}$.Let $u = x^{x}$ and $v = 2^{\sin x}$.Then $y = u - v$,so $\frac{dy}{dx} = \frac{du}{dx} - \frac{dv}{dx}$.For $u = x^{x}$,taking logarithms on both sides: $\log u = x \log x$.Differentiating with respect to $x$: $\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log x + x \cdot \frac{d}{dx}(\log x) = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.Thus,$\frac{du}{dx} = u(1 + \log x) = x^{x}(1 + \log x)$.For $v = 2^{\sin x}$,taking logarithms on both sides: $\log v = \sin x \cdot \log 2$.Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = \log 2 \cdot \frac{d}{dx}(\sin x) = \log 2 \cdot \cos x$.Thus,$\frac{dv}{dx} = v \cdot \cos x \cdot \log 2 = 2^{\sin x} \cos x \log 2$.Substituting these back,we get $\frac{dy}{dx} = x^{x}(1 + \log x) - 2^{\sin x} \cos x \log 2$.

Differentiate the function $y = x^{x} - 2^{\sin x}$ with respect to $x$.

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Assertion $(A)$: $\frac{d}{d x}\left(\frac{x^2 \sin x}{\log x}\right)=\frac{x^2 \sin x}{\log x} \left(\cot x+\frac{2}{x}-\frac{1}{x \log x}\right)$
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