If y = 2x + cot^{-1} x + log(sqrt{1 + x^2} - | vedclass

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(D) Given $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$.First,we find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = 2 - \frac{1}{1+x^2} + \frac{

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increases on $(-\infty, \infty)$

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(D) Given $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$.First,we find the derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = 2 - \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2} - x} \cdot \left( \frac{2x}{2\sqrt{1+x^2}} - 1 \right)$$= 2 - \frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2} - x} \cdot \left( \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}} \right)$$= 2 - \frac{1}{1+x^2} - \frac{1}{\sqrt{1+x^2}}$Since $x^2 \geq 0$,we have $1+x^2 \geq 1$,which implies $0 Thus,$\frac{dy}{dx} = 2 - (\frac{1}{1+x^2} + \frac{1}{\sqrt{1+x^2}}) \geq 2 - (1 + 1) = 0$.Since $\frac{dy}{dx} \geq 0$ for all $x \in \mathbb{R}$,the function $y$ increases on $(-\infty, \infty)$.

If $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$,then $y$

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