Answer
$\frac{\partial w}{\partial u}=\frac{2}{5}$
$\frac{\partial w}{\partial v}=0$
Work Step by Step
$x=2e^{u}cos(v)=2$ when $u=v=0$
$\frac{dx}{du}=2e^{u}cos(v)=2$ when $u=v=0$
$\frac{dx}{dv}=-2e^{u}sin(v)=0$ when $u=v=0$
$w=\ln\sqrt{1+x^2}-tan^{-1}x=\frac{1}{2}\ln(1+x^2)-tan^{-1}x$
$\frac{dw}{dx}=\frac{x}{{1+x^2}}-\frac{1}{{1+x^2}}=\frac{1}{5}$ when $u=v=0$
$\frac{\partial w}{\partial u}=\frac{dw}{dx}\frac{dx}{du}=\frac{1}{5}\times2=\frac{2}{5}$ when $u=v=0$
$\frac{\partial w}{\partial v}=\frac{dw}{dx}\frac{dx}{dv}=\frac{1}{5}\times0=0$ when $u=v=0$
