# Derivative of Arc Length/Proof 2

## 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 | $\quad$ |

$\blacksquare$