If $A$ is a linear map, then $A(\alpha(t))'=A(\alpha'(t))$ Let $A:\mathbb R^3\to \mathbb R^3$ be a linear map. I need the following technical result to solve a problem I'm solving:
$$A(\alpha(t))'=A(\alpha'(t))$$
Where $\alpha:I\to \mathbb R^3$ is a parametric curve.
Intuitively, I know this is true, but I would like to prove it formally.
 A: Let $T$ be the derivative of $A \circ \alpha$, evaluated at a given $t \in I$.  Then $T$ is the unique linear map $\mathbb{R} \rightarrow \mathbb{R}^3$ such that 
$$\frac{||A \circ \alpha(t + h) - A \circ \alpha(t) - T(h)||}{|h|}$$
goes to zero as $h$ goes to $0$.  Let $S: \mathbb{R} \rightarrow \mathbb{R}^3$ be the linear map $S(h) = A(\alpha'(t)h)$.  Then
$$\frac{||A \circ \alpha(t + h) - A \circ \alpha(t) - S(h)||}{|h|} = \frac{||A[ \alpha(t + h) -  \alpha(t) - \alpha'(t)h]||}{|h|}$$
where $|| \cdot ||$ is a fixed norm on $\mathbb{R}^3$.
The numerator is bounded by $||A|| \cdot ||\alpha(t+h) - \alpha(t) - \alpha'(t)h||$, where $||A||$ is the norm of the linear transformation $A$, and by definition of $\alpha'(t)$,
$$\frac{||\alpha(t+h)-\alpha(t) -\alpha'(t)h||}{|h|}$$
goes to zero as $h$ goes to $0$.  So $S$ satisfies the same property as $T$.  By uniqueness, $S = T$.
Another way using the chain rule:
The chain rule says that the derivative of $A \circ \alpha$, evaluated at $t$, is the composition of the linear transformation $\alpha'(t): \mathbb{R} \rightarrow \mathbb{R}^3$ (the derivative of $\alpha$ evaluated at $t$), followed by the linear transformation $A'(\alpha(t)): \mathbb{R}^3 \rightarrow \mathbb{R}^3$ (the derivative of $A$ evaluated at $\alpha(t)$).  So
$$(A \circ \alpha)'(t) = A'(\alpha(t)) \circ \alpha'(t)$$
Since $A$ is a linear transformation, its derivative at every point of $\mathbb{R}^3$ is constant and equal to $A$.  Thus 
$$(A \circ \alpha)(t) = A \circ \alpha'(t)$$
as linear maps $\mathbb{R} \rightarrow \mathbb{R}^3$.
