Derivative of Arc Length/Proof 2

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Theorem

Let $C$ be a curve in the cartesian coordinate plane described by the equation $y = \map f x$.

Let $s$ be the length along the arc of the curve from some reference point $P$.

Then the derivative of $s$ with respect to $x$ is given by:

$\dfrac {\d s} {\d x} = \sqrt {1 + \paren {\dfrac {\d y} {\d x} }^2}$


Proof

From Continuously Differentiable Curve has Finite Arc Length, $s$ exists and is given by:

\(\displaystyle s\) \(=\) \(\displaystyle \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} }^2} \rd u\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d s} {\d x}\) \(=\) \(\displaystyle \frac {\d} {\d x} \int_P^x \sqrt {1 + \paren {\frac {\d y} {\d u} } ^2} \rd u\) $\quad$ differentiating both sides WRT $x$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sqrt {1 + \paren {\frac {\d y} {\d x} }^2}\) $\quad$ Fundamental Theorem of Calculus/First Part $\quad$

$\blacksquare$