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What is the derivative of $\sqrt{\mathbf {\dot r}\cdot \mathbf{\dot r}}$ with respect to $\mathbf {\dot r}$?

I'm asking this due to this:

$$\delta \mathrm A = \delta \int_A^B n(\mathbf r(s) )\sqrt{\frac{\mathrm d\mathbf r}{\mathrm d s}\cdot \frac{\mathrm d\mathbf r}{\mathrm ds}}~\mathrm ds$$

which on further treatment was written as $$\delta\mathbf A = \int_A^B \left[|\mathbf{\dot r}|\frac{\partial n}{\partial{\mathbf r}}~\delta \mathbf r + \left(n(\mathbf r(s))\frac{\mathbf{\dot r}}{|\mathbf{\dot r}|}\right)~\delta{\mathbf{ \dot r}}\right]~\mathrm ds\,,$$ where $\mathbf{\dot r} = \frac{\mathrm d\mathbf r}{\mathrm ds }$ ($s$ is the arc-length parameter).

Does that mean $\frac{\mathrm d\sqrt{\mathbf {\dot r}\cdot \mathbf{\dot r}}}{\mathrm d\mathbf{\dot r} }=\frac{\mathbf{\dot r}}{|\mathbf{\dot r}|} $? I'm not getting this how it is deduced. Could anyone explain me how to get the derivative?

1 Answers1

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Note that $\frac{d}{d\textbf{x}}\textbf{x}^T\textbf{x}=2\textbf{x}$. By application of chain rule $\frac{\mathrm d\sqrt{\mathbf {\dot r}\cdot \mathbf{\dot r}}}{\mathrm d\mathbf{\dot r} } = \frac{\mathrm d\sqrt{\mathbf {\dot r}\cdot \mathbf{\dot r}}}{\mathrm d\mathbf{\dot r .\dot r} }\frac{\mathrm d{\mathbf {\dot r}\cdot \mathbf{\dot r}}}{\mathrm d\mathbf{\dot r } }= \frac{1}{2 \mathbf{\sqrt{\dot r.\dot r}}} 2\mathbf{\dot r}=\frac{\mathbf{\dot r}}{|\mathbf{\dot r}|}$

user1131274
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