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Given $y=\log(\sqrt{x}+\frac{1}{\sqrt{x}})^{2}$. Using the property of logarithms, $\log a^b = b \log a$, we have $$y = 2\log(\sqrt{x}+\frac{1}{\sqrt{x}})$$ $$y = 2\log(\frac{x+1}{\sqrt{x}})$$ $$y = 2[\log(x+1) - \log(\sqrt{x})]$$ $$y = 2\log(x+1) - 2\log(x^{1/2})$$ $$y = 2\log(x+1) - \log(x)$$
Differentiating $y$ with respect to $x$, we get $$y_1 = \frac{dy}{dx} = 2\frac{1}{x+1} - \frac{1}{x}$$ $$y_1 = \frac{2x - (x+1)}{x(x+1)} = \frac{x-1}{x(x+1)}$$
Differentiating $y_1$ with respect to $x$, we get $$y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{x-1}{x(x+1)}\right) = \frac{d}{dx} \left(\frac{x-1}{x^2+x}\right)$$ Using the quotient rule, $\frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2}$, we have $$y_2 = \frac{(x^2+x)(1) - (x-1)(2x+1)}{(x(x+1))^2} = \frac{x^2+x - (2x^2+x-2x-1)}{x^2(x+1)^2}$$ $$y_2 = \frac{x^2+x - 2x^2 + x + 1}{x^2(x+1)^2} = \frac{-x^2+2x+1}{x^2(x+1)^2}$$
We need to show that $x(x+1)^{2}y_{2}+(x+1)^{2}y_{1}=2$. Substituting the values of $y_1$ and $y_2$, we have $$x(x+1)^2 \left(\frac{-x^2+2x+1}{x^2(x+1)^2}\right) + (x+1)^2 \left(\frac{x-1}{x(x+1)}\right)$$ $$= \frac{-x^2+2x+1}{x} + \frac{(x+1)(x-1)}{x} = \frac{-x^2+2x+1}{x} + \frac{x^2-1}{x}$$ $$= \frac{-x^2+2x+1+x^2-1}{x} = \frac{2x}{x} = 2$$ Thus, $x(x+1)^{2}y_{2}+(x+1)^{2}y_{1}=2$.
Final Answer: $x(x+1)^{2}y_{2}+(x+1)^{2}y_{1}=2$ is verified.
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