Derivative of a vector valued function under an orthogonal transformation

Question

Let $\alpha : \mathbb{R} \to \mathbb{R}^3$ be a space curve. I'm trying to show that its curvature, torsion, and arc length are invariant under orthogonal transformations. If $\rho: \mathbb{R}^3\to \mathbb{R}^3$ is an orthogonal transformation, I've already shown that it preserves the norm of a vector, the angle between two vectors, and the cross product of two vectors(we can assume $\rho$ has positive determinant).

From here I've shown that curvature, torsion, and arc length are preserved under $\rho$. This is easy since they all have formulas in terms of $\alpha'$, $\alpha''$, and $\alpha'''$, and I think $(\rho(\alpha))' = \rho (\alpha')$, but I haven't proven it.

So how can I prove that $(\rho(\alpha))' = \rho (\alpha')$?

Answer 1

$$\frac{d}{dt}(\rho(\alpha(t)))=d\rho(\alpha')=\rho(\alpha')$$ The first equation holds since it is the definition of a differential. The second holds since the differential of a linear function is itself.