Assuming that $x:\Bbb{R}^{+}\to\Bbb{R}^n$ be differentiable. I want to prove that
\begin{align}f:\Bbb{R}^{+}\to\Bbb{R}\end{align} \begin{align}t\mapsto f(t)=\Vert x(t)\Vert^2\end{align} is differentiable and then, compute \begin{align}\frac{d}{dt}\Vert x(t)\Vert^2\end{align} where $\Vert x\Vert$ is the Euclidean.
MY WORK
I know that the answer is: \begin{align}\frac{d}{dt}\Vert x(t)\Vert^2=2\dot{x}\cdot x\end{align} but I want to prove it. Nevertheless, I am thinking of using composition of functions and then use bilinearity. So, I have
\begin{align}k:\Bbb{R}^{+}\to\Bbb{R}^n\backslash {0}\times \Bbb{R} \end{align} \begin{align}t\mapsto (x(t),\langle x(t),x(t)\rangle)\end{align} and \begin{align}g:\Bbb{R}^n\backslash {0}\times \Bbb{R}\to \Bbb{R} \end{align} \begin{align}(x(t),\langle x(t),x(t)\rangle)\mapsto \Vert x(t)\Vert^2\end{align} Then, $f=g\circ k$ so that \begin{align}f'(x)(k)=g'\left(h(x))(h'(x)k\right)\end{align} \begin{align}=g'\left((x(t),\langle x(t),x(t)\rangle))(x'(t),2\Vert x(t)\Vert^2 x'(t)\right)\end{align} \begin{align}=g\left((x(t),2 \langle x(t),x(t)\rangle x'(t)))+g(x'(t),\langle x(t),x(t)\rangle\right)\end{align} Even with this approach of mine, I'm not seeing a way-through! Any help please?
I don’t understand why you use a complex map like $k$.
You can write $f=f_2\circ f_1$, where $f_1(t)=x(t)$ and $f_2: x \mapsto \Vert x \Vert^2 =\langle x,x \rangle$.
The Fréchet derivative of $f_2$ at $x$ is $f_2^\prime(x).u = 2 \langle x,u\rangle$ as $$\langle x+u,x+u\rangle = \langle x ,x \rangle+2\langle x,u\rangle +\langle u ,u \rangle$$
and $\lim\limits_{u \to 0} \dfrac{\langle u, u \rangle}{\Vert u \Vert}=0$.
Then applying the chain rule you get the result $f^\prime(t)= 2\langle x(t),x^\prime(t)\rangle$.