Derivative in normed space

Question

I'm studying the Differential Calculus Functions of several variables. Let $f:A\subset\mathbb{R}\to\mathbb{R}^n$ is differentiable with $||f(t)||>0$ for all $t\in A$. Prove that: $$u(t)=\dfrac{f(t)}{||f(t)||}$$ is differentiable and find $<u(t),u'(t)>$

I know the way to check $u(t)$ is differentiable or not that is calculating this limit: $$\lim\limits_{h\to0}{\dfrac{||u(t+h)-u(t)||}{|h|}};$$ And what I got next is: $$\lim\limits_{h\to0}{\dfrac{\bigg|\bigg|\dfrac{f(t+h)}{||f(t+h)||}-\dfrac{f(t)}{||f(t)||}\bigg|\bigg|}{|h|}}$$ I found out that: $$\dfrac{\bigg|\bigg|\dfrac{f(t+h)}{||f(t+h)||}-\dfrac{f(t)}{||f(t)||}\bigg|\bigg|}{|h|}\le\dfrac{1}{|h|}\bigg|\bigg|\dfrac{f(t+h)-f(t)}{||f(t+h)||}+\left(\dfrac{1}{||f(t+h)||}-\dfrac{1}{||f(t)||}\right)f(t)\bigg|\bigg|$$ $$\le\dfrac{1}{||f(t+h)||}\dfrac{||f(t+h)-f(t)||}{|h|}+\dfrac{1}{|h|}\bigg|\dfrac{1}{||f(t+h)||}-\dfrac{1}{||f(t)||}\bigg|||f(t)||;$$ Let $|h|\to0$ then: $$\lim\limits_{h\to0}{\dfrac{||u(t+h)-u(t)||}{|h|}}\le2\dfrac{||f'(t)||}{||f(t)||}$$ but I see it's not useful to my problem. Please help me.

Answer 1

The function $u$ is differentiable as a ratio of differentiable functions $f$ and $|f\|.$ We have $1=\|u(t)\|^2=\langle u(t),u(t)\rangle.$ Thus $$ 0={d\over dt}\langle u(t),u(t)\rangle=2\langle u(t),u'(t)\rangle.$$ Remark There is no need for calculating $u'.$

Answer 2

The $i$-th component of $u(t)$ is $$u_i(t)=(f_1^2(t)+\ldots+f_n^2(t))^{-\frac{1}{2}}f_i(t)$$

Using the product rule, you find $$ \begin{aligned} u_i'(t)&=-\frac{1}{2}(f_1^2(t)+\ldots+f_n^2(t))^{-\frac{3}{2}} 2(f_1(t)f_1'(t)+\ldots+f_n(t)f_n'(t))f_i(t)\\ &\quad+(f_1^2(t)+\ldots+f_n^2(t))^{-\frac{1}{2}}f_i'(t)\\ &=-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^3}f_i(t)+\frac{1}{\lVert f(t)\rVert}f_i'(t), \end{aligned}$$

such that

$$u'(t)=-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^3}f(t)+\frac{f'(t)}{\lVert f(t)\rVert}.$$

This yields

$$ \begin{aligned} \langle u(t),u'(t)\rangle &= \langle \frac{f(t)}{\lVert f(t) \rVert},-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^3}f(t)+\frac{f'(t)}{\lVert f(t)\rVert}\rangle\\ &= \langle \frac{f(t)}{\lVert f(t) \rVert},-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^3}f(t)\rangle + \langle \frac{f(t)}{\lVert f(t) \rVert},\frac{f'(t)}{\lVert f(t)\rVert}\rangle\\ &=-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^4}\langle f(t),f(t)\rangle+\frac{1}{\lVert f(t) \rVert^2}\langle f(t),f'(t)\rangle\\ &=-\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^2}+\frac{\langle f(t),f'(t)\rangle}{\lVert f(t)\rVert^2}\\ &=0. \end{aligned}$$