Show that $G$ is differentiable for all $t$, and find $G'(0)$

Question

Let $ \vec c : \mathbb{R} \to \mathbb{R^3}$, be a differentiable function on $\mathbb{R}$, with $\vec c(0)$ = (1, 2, 3) and $D\vec c(0) =$ $$\left [\begin{matrix} 1&\\-1&\\2 \end{matrix}\right] $$

Consider $G : \mathbb{R} \to \mathbb{R}$ defined by $G(t) = ||\vec c(t)||^2$. Show that G is differentiable for all $t$, and find $DG(0)$ (or $G'0$; same quantity).

I have no idea how to start this question. I am confused as to what the first part of the question is even telling me.

Any guidance is appreciated!

Answer 1

To prove that $G$ is differentiable, it suffices to prove that $n(x) = \|x\|^2$ is differentiable. Note that \begin{align*} n(x+h) &= \|x+h\|^2\\ &= \def\<#1>{\left<#1\right>}\\\ &= \ + 2\ + \\\ &= n(x) + 2\ + o(\|h\|) \end{align*} Hence, $n$ is differentiable with $Dn(x)h = 2\$. Therefore, by the chain rule, $G$ is differentiable, and $$ DG(t) = Dn\bigl(c(t)\bigr)Dc(t) = 2\ $$ At $t = 0$, we have $$ DG(0) = 2\ = 10. $$