Question

$\frac{d}{dx}\left[{\log }_e\left\{(e^x+2)+\sqrt{e^{2x}+4e^x+5}\right\}\right]=$

• A. $\frac{1}{\sqrt{e^{2x}+4e^x+5}}$
• B. $\frac{e^x}{\sqrt{e^{2x}+4e^x+5}}$
• C. $\frac{e^x}{\sqrt{e^{2x}+4e^x+3}}$
• D. $\frac{-e^x}{\sqrt{e^{2x}+4e^x+3}}$

Answer 1

The correct option is B $\frac{e^x}{\sqrt{e^{2x}+4e^x+5}}$

$\frac{d}{dx}\left[{\log }_e\left\{\right(e^x+2)+\sqrt{e^{2x}+4x+5}\}\right]$$=\frac{d}{dx}\left[{\log }_e\left\{\right(e^x+2)+\sqrt{(e^2+2^2+1}\}\right]$Let $z=e^x+2\Rightarrow e^x=z-2$$\Rightarrow dz=e^{x}dx\Rightarrow dx=\frac{1}{(z-2)}dz$$\Rightarrow \frac{d}{dx}=(z-2)\frac{d}{dz}$$\Rightarrow \frac{d}{dx}\left[{\log }_e\{(e^x+2)+\sqrt{e^{2x}+4x+5}\}\right]=(z-2)\frac{d}{dz}\left[{\log }_e\{z+\sqrt{z^2+1}\}\right]$Now $\frac{d}{dz}\left[{\log }_e\{z+\sqrt{z^2+1}\}\right]=\frac{1\times 1+\frac{2z}{2\sqrt{z^2+1}}}{2+\sqrt{z^2+1}}$$=\frac{z+\sqrt{z^2+1}}{(z+\sqrt{z^2+1})\left(\sqrt{z^2+1}\right)}$$=\frac{1}{\sqrt{z^2+1}}$$=\frac{1}{\sqrt{e^{2x}+4e^x+5}}$$\therefore (z-2)\frac{d}{dz}\left[{\log }_e\{z+\sqrt{z^2+1}\}\right]=\frac{e^x}{\sqrt{e^{2x}+4e^x+5}}$$\therefore \frac{d}{dx}\left[{\log }_e\left\{\right(e^x+2)+\sqrt{e^{2x}+4e^x+5}\}\right]=\frac{e^x}{\sqrt{e^{2x+4e^x+5}}}$