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

If $a>0$ and $f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2 x}$, then $f^{\prime}(0)=$
  1. $a^{a+1}$
  2. $a^{a+1}\left\{\frac{1-a^2}{a}+2 \log a\right\}$
  3. $2 \log a$
  4. $a^{a+1}\left\{\frac{(1+a)^2}{a-2 \log a}\right\}$

Solution

Given function, for $a>0$ $ \begin{aligned} & f(x)=\left(\frac{a+x}{1+x}\right)^{a+1+2 x} \Rightarrow f(0)=a^{a+1} \\ \Rightarrow \quad & \log f(x)=(a+1+2 x) \log \left(\frac{a+x}{1+x}\right) \end{aligned} $ On differentiating both side with respect to ' $x$ ', we are getting $ \begin{aligned} & \frac{f^{\prime}(x)}{f(x)}=2 \log \left(\frac{a+x}{1+x}\right)+(a+1+2 x)\left\lceil\frac{1}{a+x}-\frac{1}{1+x}\right\rfloor \\ & \text { So, } f^{\prime}(0)=f(0)\left[2 \log (a)+(a+1)\left(\frac{1}{a}-1\right)\right] \\ & \because \quad f(0)=a^{a+1} \end{aligned} $ So, $f^{\prime}(0)=a^{a+1}\left\{\frac{1-a^2}{a}+2 \log a\right\}$

Asked in: AP EAMCET 2018 (23 Apr Shift 2)

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