If y = left(frac{x^{2}}{x+1}right)^{x} and fr | vedclass

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(A) Given $y = \left(\frac{x^{2}}{x+1}\right)^{x}$.Taking natural logarithm on both sides:$\log y = x \log \left(\frac{x^{2}}{x+1}\right) = x [\log(x^

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$\frac{x(x+2)}{x+1}$

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(A) Given $y = \left(\frac{x^{2}}{x+1}\right)^{x}$.Taking natural logarithm on both sides:$\log y = x \log \left(\frac{x^{2}}{x+1}\right) = x [\log(x^{2}) - \log(x+1)] = x [2 \log x - \log(x+1)]$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [x (2 \log x - \log(x+1))]$.Using the product rule:$\frac{1}{y} \frac{dy}{dx} = 1 \cdot [2 \log x - \log(x+1)] + x \left[ \frac{2}{x} - \frac{1}{x+1} \right]$.$\frac{1}{y} \frac{dy}{dx} = \log \left(\frac{x^{2}}{x+1}\right) + x \left[ \frac{2(x+1) - x}{x(x+1)} \right]$.$\frac{1}{y} \frac{dy}{dx} = \log \left(\frac{x^{2}}{x+1}\right) + x \left[ \frac{2x + 2 - x}{x(x+1)} \right] = \log \left(\frac{x^{2}}{x+1}\right) + \frac{x+2}{x+1}$.Thus,$\frac{dy}{dx} = y \left[ \frac{x+2}{x+1} + \log \left(\frac{x^{2}}{x+1}\right) \right]$.Comparing this with the given form $\frac{dy}{dx} = y [g(x) + \log \left(\frac{x^{2}}{x+1}\right)]$,we get $g(x) = \frac{x+2}{x+1}$.

If $y = \left(\frac{x^{2}}{x+1}\right)^{x}$ and $\frac{dy}{dx} = y \left[g(x) + \log \left(\frac{x^{2}}{x+1}\right)\right]$,then $g(x) =$

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