Choose the correct alternative in the following:

$\frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left\{\mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)^{\frac{2}{4}}\right\}\right]$

Let $u=\frac{x-2}{x+2} \Rightarrow \frac{d u}{d x}=\frac{1 \cdot(x+2)-(x-2) \cdot 1}{(x+2)^{2}}=\frac{4}{(x+2)^{2}} \cdots(1)$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left\{\mathrm{e}^{\mathrm{x}}(\mathrm{u})^{\frac{3}{4}}\right\}\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\log \mathrm{e}^{\mathrm{x}}+\log (\mathrm{u})^{\frac{3}{4}}\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x} \cdot \log \mathrm{e}+\frac{3}{4} \log (\mathrm{u})\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{x}+\frac{3}{4} \log (\mathrm{u})\right] \because \log \mathrm{e}=1$

$\Rightarrow 1+\frac{3}{4} \cdot \frac{1}{\mathrm{u}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\Rightarrow 1+\frac{3}{4} \cdot \frac{(\mathrm{x}+2)}{\mathrm{x}-2} \cdot \frac{4}{(\mathrm{x}+2)^{2}}-$ From (1)

$\Rightarrow 1+\frac{3}{\left(\mathrm{x}^{2}-2^{2}\right)}$

$\Rightarrow \frac{\left(\mathrm{x}^{2}-4\right)+3}{\left(\mathrm{x}^{2}-4\right)}$

$\therefore \frac{\mathrm{d}}{\mathrm{dx}}\left[\log \left\{\mathrm{e}^{\mathrm{x}}\left(\frac{\mathrm{x}-2}{\mathrm{x}+2}\right)^{\frac{3}{4}}\right\}\right]=\frac{\mathrm{x}^{2}-1}{\mathrm{x}^{2}-4}=(\mathrm{A})$