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(D) Given the equation: ${x^{2x}} - 2{x^x}\cot y - 1 = 0$  ........$(i)$At $x = 1$,we have:$1^{2(1)} - 2(1^1)\cot y - 1 = 0$$1 - 2\cot y - 1 = 0$$\Rig

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$-1$

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(D) Given the equation: ${x^{2x}} - 2{x^x}\cot y - 1 = 0$  ........$(i)$At $x = 1$,we have:$1^{2(1)} - 2(1^1)\cot y - 1 = 0$$1 - 2\cot y - 1 = 0$$\Rightarrow \cot y = 0 \quad 	herefore \quad y = \frac{\pi}{2}$Differentiating $(i)$ with respect to $x$:$\frac{d}{dx}(x^{2x}) - 2 \frac{d}{dx}(x^x \cot y) = 0$$x^{2x} \cdot \frac{d}{dx}(2x \ln x) - 2 \left[ x^x \ln x \cdot \cot y + x^x \cdot (-\csc^2 y) \frac{dy}{dx} + \cot y \cdot x^x(1 + \ln x) ight] = 0$Using $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$:$2x^{2x}(1 + \ln x) - 2 \left[ x^x(1 + \ln x) \cot y - x^x \csc^2 y \frac{dy}{dx} ight] = 0$At $P(1, \frac{\pi}{2})$,we have $x = 1, y = \frac{\pi}{2}, \cot y = 0, \csc^2 y = 1, \ln 1 = 0$:$2(1)^{2}(1 + 0) - 2 \left[ 1(1 + 0)(0) - 1(1) \left(\frac{dy}{dx}ight)_{P} ight] = 0$$2 - 2 \left[ -\left(\frac{dy}{dx}ight)_{P} ight] = 0$$2 + 2 \left(\frac{dy}{dx}ight)_{P} = 0$$\left(\frac{dy}{dx}ight)_{P} = -1$

Let $y$ be an implicit function of $x$ defined by ${x^{2x}} - 2{x^x}\cot y - 1 = 0$. Then $y'(1)$ equals

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