If a0 and f(x)=left(frac{a+x}{1+x}right)^{a+ | vedclass

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(B) Given the function $f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x}$ for $a>0$. Taking the natural logarithm on both sides: $\log f(x) = (a+1+2x) \log

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$a^{a+1}\left\{\frac{1-a^2}{a}+2 \log a\right\}$

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(B) Given the function $f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x}$ for $a>0$. Taking the natural logarithm on both sides: $\log f(x) = (a+1+2x) \log \left(\frac{a+x}{1+x}\right)$. Differentiating both sides with respect to $x$: $\frac{f^{\prime}(x)}{f(x)} = 2 \log \left(\frac{a+x}{1+x}\right) + (a+1+2x) \left(\frac{1}{a+x} - \frac{1}{1+x}\right)$. Evaluating at $x=0$: $\frac{f^{\prime}(0)}{f(0)} = 2 \log \left(\frac{a}{1}\right) + (a+1) \left(\frac{1}{a} - 1\right)$. $\frac{f^{\prime}(0)}{f(0)} = 2 \log a + (a+1) \left(\frac{1-a}{a}\right)$. Since $f(0) = a^{a+1}$,we have: $f^{\prime}(0) = a^{a+1} \left[ 2 \log a + \frac{(a+1)(1-a)}{a} \right]$. $f^{\prime}(0) = a^{a+1} \left[ 2 \log a + \frac{1-a^2}{a} \right]$. Thus,the correct option is $B$.

If $a>0$ and $f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2x}$,then $f^{\prime}(0)=$

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