frac{d}{d x} [x^{sin x}+(sin x)^x]= | vedclass

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(A) Let $y = x^{\sin x} + (\sin x)^x$. Let $U = x^{\sin x}$ and $V = (\sin x)^x$. Then $\frac{dy}{dx} = \frac{dU}{dx} + \frac{dV}{dx}$. For $U = x^{\s

Vedclass · Accepted Answer

$x^{\sin x} [\frac{\sin x}{x}+\cos x \log x]+(\sin x)^x [\log \sin x+x \cot x]$

Vedclass · Answer

(A) Let $y = x^{\sin x} + (\sin x)^x$. Let $U = x^{\sin x}$ and $V = (\sin x)^x$. Then $\frac{dy}{dx} = \frac{dU}{dx} + \frac{dV}{dx}$. For $U = x^{\sin x}$,taking $\log$ on both sides: $\log U = \sin x \log x$. Differentiating with respect to $x$: $\frac{1}{U} \frac{dU}{dx} = \cos x \log x + \sin x \cdot \frac{1}{x} = \frac{\sin x}{x} + \cos x \log x$. So,$\frac{dU}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x]$. For $V = (\sin x)^x$,taking $\log$ on both sides: $\log V = x \log(\sin x)$. Differentiating with respect to $x$: $\frac{1}{V} \frac{dV}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x = \log(\sin x) + x \cot x$. So,$\frac{dV}{dx} = (\sin x)^x [\log(\sin x) + x \cot x]$. Thus,$\frac{dy}{dx} = x^{\sin x} [\frac{\sin x}{x} + \cos x \log x] + (\sin x)^x [\log(\sin x) + x \cot x]$.

$\frac{d}{d x} [x^{\sin x}+(\sin x)^x]=$

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