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Let $y = x^{\cot x} + \frac{2x^{2}-3}{2x^{2}-x+2}$. We need to find $\frac{dy}{dx}$. Let $u = x^{\cot x}$ and $v = \frac{2x^{2}-3}{2x^{2}-x+2}$. Then $y = u + v$, so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$.
Taking log on both sides: $\ln u = \cot x \ln x$. Differentiating with respect to $x$: $$\frac{1}{u} \frac{du}{dx} = (-\csc^2 x) \ln x + (\cot x) \frac{1}{x}$$ $$\frac{du}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \ln x \right)$$
Using the quotient rule $\frac{d}{dx}(\frac{f}{g}) = \frac{f'g - fg'}{g^2}$: $$f' = 4x, g' = 4x - 1$$ $$\frac{dv}{dx} = \frac{(4x)(2x^2 - x + 2) - (2x^2 - 3)(4x - 1)}{(2x^2 - x + 2)^2}$$ Simplifying the numerator: $(8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3) = -2x^2 + 20x - 3$$ $$\frac{dv}{dx} = \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2}$$
The final derivative is: $$\frac{dy}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \ln x \right) + \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2}$$
Final Answer: x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \ln x \right) + \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2}
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