Let $\gamma:(\alpha,\beta)\rightarrow \mathbb{R}^n$ a smooth regular curve. To define the curvature of $\gamma$ at the parameter value $t \in (\alpha,\beta)$, we take a smooth unit-speed reparametrization $\delta:(a,b)\rightarrow \mathbb{R}^n$ (with $\gamma=\delta \circ \phi$, where $\phi:(\alpha,\beta)\rightarrow (a,b)$ is a diffeomorphism) and take the quantity $||\ddot \delta(\phi(t))||$ to be such curvature.
How can I show that this definition doesn't depend on the reparametrization? That is, how can I show that if I take another smooth unit speed curve $\delta_1:(a_1,b_1)\rightarrow \mathbb{R}^n$ and another diffeomorphism $\phi_1:(\alpha,\beta)\rightarrow (a_1,b_1)$ such that $\gamma = \delta_1 \circ \phi_1$, I have $||\ddot\delta(\phi(t))||=||\ddot\delta_1(\phi_1(t)||?$