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We can use the identity $\mathrm d\mathbf x=\mathbf F\,\mathrm d\mathbf X$ relating the line elements in the reference $(\mathrm d\mathbf X)$ and current $(\mathrm d\mathbf x)$ configurations to prove that $$\dot{\overline{\mathrm d\mathbf x}}=\mathbf L\,\mathrm d\mathbf x,$$ where $\mathbf L$ is the velocity gradient at the current location of the element.

My proof: \begin{equation} \dot{\overline{\mathrm d\mathbf x}}=\dot{\overline{\mathbf F\,\mathrm d\mathbf X}} =\mathbf{\dot{F}}\,\mathrm d\mathbf X =\mathbf{LF}\,\mathrm d\mathbf X =\mathbf L\,\mathrm d\mathbf x. \end{equation}

But I am not sure why we do not have $\mathbf{\dot{F}}\,\mathrm d\mathbf X+\mathbf F\,\mathrm d\dot{\mathbf{X}}$ in the third equation of the proof.

Is it because line elements are not differentiable or the term $\mathbf F\,\mathrm d\dot{\mathbf{X}}$ is zero?

I would appreciate any help or hint. Thank you.

EDIT: the superposed dot represents the material time derivative, $\mathbf F$ is the deformation gradient tensor and in the proof I have used the result $\dot{\mathbf{F}}=\mathbf{LF}$.

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    The answer is probably that [$\textrm{d}^2=0$](http://math.stackexchange.com/questions/1425421/coordinate-free-proof-of-d2-0), but just out of curiousity, what does the overline mean? – 2017-02-12
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    @Bye_World The overline: I am trying to make clear that the material derivative is taken of the whole expression. – 2017-02-12
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    I see. Well I've never studied continuum mechanics and so I'm not sure of all the notation here, but I'd still bet what you're looking for is the $\textrm{d}^2=0$ identity. – 2017-02-12
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    To say $d\mathbf{X}$ denotes a "reference line element" suggests a constant one-form, akin to the Cartesian differentials $dx$, $dy$, and $dz$. (Offhand, I'd guess that $d\mathbf{X}$ _is_ the vector-valued one-form $(dx, dy, dz)$...?) If that's right, there's no term for $\dot{d\mathbf{X}}$ because $d\mathbf{X}$ is constant. – 2017-02-12

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