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# Why do differentials have a linear relationship?

2017-12-14 05:23:16

I recently learned that $$\mathrm dz = \frac{\partial f}{\partial x} \mathrm dx + \frac{\partial f}{\partial y} \mathrm dy$$

How is this result derived? If you asked me, I would have said something like $$\big(\mathrm dz\big)^2 = \big(\frac{\partial f}{\partial x} \mathrm dx\big)^2 + \big(\frac{\partial f}{\partial y} \mathrm dy \big)^2$$ (assuming that taking the square of differentials is a valid operation.)

Doesn't the equation [in its correct form] imply that the change in $z$ is linearly dependent on $x$ and $y$? Why isn't this tantamount to saying that $||\vec v|| = v_x + v_y$ for some 2D vector?