Answer
$du=\frac{x}{\sqrt{x^2+3y^2}} dx + \frac{3y}{\sqrt{x^2+3y^2}} dy$
Work Step by Step
The differential is defined as:
$dw=\frac{\partial w}{\partial x} dx + \frac{\partial w}{\partial y} dy + \frac{\partial w}{\partial z} dz$
Given the function $u=\sqrt{x^2+3y^2}$ . Thus, we find the partial derivatives w.r.t $x$ and $y$:
$f_x=\frac{1}{2}(x^2+3y^2)^{-\frac{1}{2}} \cdot 2x=\frac{x}{\sqrt{x^2+3y^2}}$
$f_y=\frac{3y}{\sqrt{x^2+3y^2}}$
Thus, the differential:
$du=\frac{x}{\sqrt{x^2+3y^2}} dx + \frac{3y}{\sqrt{x^2+3y^2}} dy$
