Answer
$f_{x}(x, y)=\displaystyle \frac{1}{x+\sqrt{x^{2}+y^{2}}}(1+\frac{x}{\sqrt{x^{2}+y^{2}}})$
$f_{x}(3,4)=\displaystyle \frac{1}{5}$.
Work Step by Step
$f(x, y) =\ln(x+\sqrt{x^{2}+y^{2}})=\ln(x+(x^{2}+y^{2})^{1/2})$
Treat y as constant to calculate $f_{x}(x, y)$. We need the chain rule.
$f_{x}(x, y)=\displaystyle \frac{1}{x+\sqrt{x^{2}+y^{2}}}\cdot\frac{\partial}{\partial x}[x+(x^{2}+y^{2})^{1/2}]$
$=\displaystyle \frac{1}{x+\sqrt{x^{2}+y^{2}}}[1+\frac{1}{2}(x^{2}+y^{2})^{-1/2}\cdot(2x)]$
$=\displaystyle \frac{1}{x+\sqrt{x^{2}+y^{2}}}(1+\frac{x}{\sqrt{x^{2}+y^{2}}})$
$f_{x}(3,4)=\displaystyle \frac{1}{3+\sqrt{3^{2}+4^{2}}}(1+\frac{3}{\sqrt{3^{2}+4^{2}}})$
$=\displaystyle \frac{1}{8}(1+\frac{3}{5})$
$=\displaystyle \frac{1}{5}$.
