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

Differentiating with respect to $x$

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

$=\frac{1}{x+\sqrt{x^{2} -a^{2}}} \ \cdotp \ \frac{d}{dx}\left( x+\ \sqrt{x^{2} -a^{2}} \ \right)$

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

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

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

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

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

$=\frac{1}{\sqrt{x^{2} -a^{2}}}$