If y=x^{sin x}+(sin x)^x,then frac{d y}{d x} | vedclass

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(C) Given,$y=x^{\sin x}+(\sin x)^x$. Let $u=x^{\sin x}$ and $v=(\sin x)^x$. Then $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$. For $u=x^{\sin x}$,t

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(C) Given,$y=x^{\sin x}+(\sin x)^x$. 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 $x$: $\frac{1}{u} \frac{du}{dx} = \sin x \cdot \frac{1}{x} + \cos x \log x$. So,$\frac{du}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \log x \right)$. For $v=(\sin x)^x$,taking log on both sides: $\log v = x \log(\sin x)$. Differentiating with respect to $x$: $\frac{1}{v} \frac{dv}{dx} = x \cdot \frac{\cos x}{\sin x} + \log(\sin x) = x \cot x + \log(\sin x)$. So,$\frac{dv}{dx} = (\sin x)^x (x \cot x + \log(\sin x))$. At $x = \frac{\pi}{2}$: $\frac{du}{dx} = (\frac{\pi}{2})^{\sin(\pi/2)} (\frac{\sin(\pi/2)}{\pi/2} + \cos(\pi/2) \log(\pi/2)) = (\frac{\pi}{2})^1 (\frac{1}{\pi/2} + 0) = \frac{\pi}{2} \cdot \frac{2}{\pi} = 1$. $\frac{dv}{dx} = (\sin(\pi/2))^{\pi/2} (\frac{\pi}{2} \cot(\pi/2) + \log(\sin(\pi/2))) = (1)^{\pi/2} (\frac{\pi}{2} \cdot 0 + \log(1)) = 1 \cdot (0 + 0) = 0$. Therefore,$\frac{dy}{dx} = 1 + 0 = 1$.

If $y=x^{\sin x}+(\sin x)^x$,then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is

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