ddxloge⁡ex+2+e2x+4ex+5=

ddxloge⁡ex+2+e2x+4ex+5=ddxlogex+2+1+ex+22ddxsinh-1ex+2=ex1+ex+22=exe2x+4ex+5

ddxloge⁡ex+2+e2x+4ex+5=

ddxloge⁡ex+2+e2x+4ex+5=ddxlogex+2+1+ex+22ddxsinh-1ex+2=ex1+ex+22=exe2x+4ex+5

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$\frac{d}{d x} [\log_{e} \{(e^{x} + 2) + \sqrt{e^{2 x} + 4 e^{x} + 5}\}] =$

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$\frac{1}{\sqrt{e^{2 x} + 4 e^{x} + 5}}$

$\frac{e^{x}}{\sqrt{e^{2 x} + 4 e^{x} + 5}}$

$\frac{e^{x}}{\sqrt{e^{2 x} + 4 e^{x} + 3}}$

$\frac{- e^{x}}{\sqrt{e^{2 x} + 4 e^{x} + 3}}$

answer is B.

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Detailed Solution

$\frac{d}{d x} [\log_{e} \{(e^{x} + 2) + \sqrt{e^{2 x} + 4 e^{x} + 5}\}] =$

$\frac{d}{d x} \log ((e^{x} + 2) + \sqrt{1 + {(e^{x} + 2)}^{2}})$

$\frac{d}{d x} \sin h^{- 1} (e^{x} + 2)$

$= \frac{e^{x}}{\sqrt{1 + {(e^{x} + 2)}^{2}}} = \frac{e^{x}}{\sqrt{e^{2 x} + 4 e^{x} + 5}}$

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$\frac{d}{d x} [\log_{e} \{(e^{x} + 2) + \sqrt{e^{2 x} + 4 e^{x} + 5}\}] =$