$\log \left( {x + \sqrt {{x^2} + a} } \right)$

Let $y = \log \left( {x + \sqrt {{x^2} + a} } \right)$

therefore,$\frac{{dy}}{{dx}} = \frac{d}{{dx}}\log \left( {x + \sqrt {{x^2} + a} } \right)$

$= \frac{1}{{\left( {x + \sqrt {{x^2} + a} } \right)}} \cdot \frac{d}{{dx}}\left[ {x + \sqrt {{x^2} + a} } \right]$

$= \frac{1}{{\left( {x + \sqrt {{x^2} + a} } \right)}}\left[ {1 + \frac{1}{2}{{\left( {{x^2} + a} \right)}^{ - 1/2}} \cdot 2x} \right]$

$= \frac{1}{{\left( {x + \sqrt {{x^2} + a} } \right)}} \cdot \left( {1 + \frac{x}{{\sqrt {{x^2} + a} }}} \right)$

$= \frac{{\left( {\sqrt {{x^2} + a} + x} \right)}}{{\left( {x + \sqrt {{x^2} + a} } \right)\left( {\sqrt {{x^2} + a} } \right)}} = \frac{1}{{\left( {\sqrt {\left. {{x^2} + a} \right)} } \right.}}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{{\left( {\sqrt {\left. {{x^2} + a} \right)} } \right.}}$