From the definition $\mathrm{\κ}\=∥\mathbf{T}\prime \left(s\right)∥$, obtain $\mathrm{\κ}\=∥\frac{d\mathbf{T}}{\mathrm{dp}}\frac{\mathrm{dp}}{\mathrm{ds}}∥$ = $\frac{\∥\mathbf{T}\prime \left(p\right)\∥}{\left|\mathrm{\ρ}\right|}$ by the chain rule.

Of course, the astute reader realizes that the prime is being used to designate differentiation with respect to both $s$ and $t$. However, the explicit display of the argument of T clarifies the differentiation variable.

In the following analysis of the formula $\mathrm{\kappa }\=∥\mathbf{R}\prime \left(p\right)\times \mathbf{R}″\left(p\right)∥\/{\mathrm{\rho }}^3$, the prime will represent differentiation with respect to $p$.

Since $\mathbf{R}\prime \=\mathrm{\ρ} \mathbf{T}$, $\mathbf{R}″\=\mathrm{\ρ}\prime \mathbf{T}\+\mathrm{\ρ} \mathbf{T}\prime$. Then,

where $\mathbf{T}\times \mathbf{T}\=\mathbf{0}$ because T is collinear with itself. Next, obtain

where $\parallel \mathbf{T}\times \mathbf{T}\prime \parallel$ = $∥\mathbf{T}∥$ $\parallel \mathbf{T}\prime \parallel$ because $\mathbf{T}$ and $\mathbf{T}\prime$ are orthogonal, and where $∥\mathbf{T}∥\=1$ because $\mathbf{T}$ is a unit vector. Finally, then,

$\frac{\parallel \mathbf{R}\prime \times \mathbf{R}″\parallel }{{\mathrm{\ρ}}^3}\=\frac{{\mathrm{\ρ}}^2 \parallel \mathbf{T}\prime \parallel }{{\mathrm{\ρ}}^3}$ = $\frac{\parallel \mathbf{T}\prime \parallel }{\mathrm{\ρ}}$