# What is the derivative of kinetic energy with respect to velocity?

It's the linear momentum $p = m v$.

The kinetic energy of a particle is defined as $K = \frac{1}{2} m {v}^{2}$.

It's derivative with respect to the the velocity $v$ is:

$\frac{\mathrm{dK}}{\mathrm{dv}} = \frac{d}{\mathrm{dv}} \left[\frac{1}{2} m {v}^{2}\right]$

Since the mass $m$ does not depend on the velocity and the factor $\frac{1}{2}$ is constant, the linear property of the derivative gives us:

$\frac{d}{\mathrm{dv}} \left[\frac{1}{2} m {v}^{2}\right] = \frac{1}{2} m \frac{d}{\mathrm{dv}} \left[{v}^{2}\right]$

Knowing the derivative of a power function $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$ gives us the result:

$\frac{\mathrm{dK}}{\mathrm{dv}} = \frac{1}{2} m 2 v = m v = p$

This answer is valid if we consider the classical case. Taking into account relativistic effects gives us the same result, but the derivation is more complicated.