For instance, I start with this relation:

$$ s^2=x^2+y^2 $$

Taking the total derivative on each side, I get:

$$ 2sds=2xdx+2ydy $$

Can I take the total derivative a second like this:

$$ d[sds]=d[xdx]+d[ydy]\\ dsds+sd[ds]=dxdx+xd[dx]+dydy+yd[dy]\\ (ds)^2=(dx)^2+(dy)^2 $$

where $d[d[s]]=0$.


1 Answer 1


If you are treating differentials as algebraic units, the second differential is not equal to zero. You can think of it as a second-order infinitesimal, but it is not zero. It can be represented as you did $\text{d}(\text{d}x)$, but it is more often represented as $d^2(x)$.

The total second differential is what you had before trying to simplify:

$$(\text{d}s)^2 + s\,\text{d}^2s = (\text{d}x)^2 + x\,\text{d}^2x + (\text{d}y)^2 + y\,\text{d}^2y$$

For more information on algebraically-manipulable higher-order differentials, see my paper "Extending the Algebraic Manipulability of Differentials".


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