Limits

$\begin{aligned} x^{2x} - 2x^x \cot y - 1 & = 0 & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ 2(\ln x + 1)x^{2x} - 2(\ln x +1)x^x \cot y + 2x^x \csc^2 y \frac {dy}{dx} & = 0 \\ 2(\ln 1 + 1) - 2(\ln 1 +1) \cot y(1) + 2 \csc^2 y (1) \frac {dy}{dx}\bigg|_{x=1} & = 0 & \small \color{#3D99F6} \text{See note: }\cot y(1) = 0, \ \csc y(1) = 1 \\ 2 + 2\frac {dy}{dx} \bigg|_{x=1} & = 0 \\ \implies \frac {dy}{dx} \bigg|_{x=1} & = \boxed{-1} \end{aligned}$

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Note:

$\begin{aligned} x^{2x} - 2x^x \cot y - 1 & = 0 \\ 1^2 - 2(1^1) \cot y(1) - 1 & = 0 \\ \implies \cot y(1) & = 0 \end{aligned}$

$x^{2x}-2x^x\cot{y}-1=0 \\ (x^x)^2-2(x^x)(\cot{y})+\cot^2{y}-1-\cot^2{y}=0 \\ (x^x-\cot{y})^2=1+\cot^2{y} \\ x^x-\cot{y}=\csc{y} \\ x^x=\cot{y}+\csc{y} \\ x^x(1+\ln{x})=-\csc^2{y}\times y^{'}-\csc{y}\cot{y}\times y^{'} \\ x^x(1+\ln{x})=-\csc{y}(\csc{y}+\cot{y})y^{'} \\ y^{'}=-\dfrac{x^x(1+\ln{x})}{x^x\csc{y}} \\ y^{'}(1)=-\dfrac{1^1(1+\ln{1})}{\csc{(y(1))}} \implies -\sin{(y(1))} \\ \text{From our Original Equation , } 1^{2}-2 \times 1^1\cot{y(1)}-1=0 \implies y(1)=\dfrac{\pi}{2} \\ y^{'}(1)=-\sin{(y(1))} \implies \color{#3D99F6}{\boxed{-1}}$

$-\cot y = \displaystyle\frac{1-x^{2x}}{2x^x}$

$-\tan y= \displaystyle\frac{2 x^x}{1-x^{2x}}$

$y = -\displaystyle\tan^{-1}(\frac{2x^x}{1-x^{2x}})$

$y =-2\tan^{-1}(x^x)$

$\displaystyle\frac{dy}{dx}=-\frac{2x^x(\ln x+1)}{(1+x^{2x})}$

$\displaystyle\left (\frac{dy}{dx}\right)_{x=1}=-1$