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I want to prove that $f$ is differentiable.
I use the principle of differentiability that says that $f$ is differentiable if there exists an A such that:

$$\lim \cfrac{\|f(t)-f(\xi)-A(t-\xi)\|}{\|t-\xi\|}=0, t\rightarrow \xi$$

I know that $A$ is the derivative of $f$ thus $A=(-\sin t, \cos t)$. But when I substitute this in the equation I get that:

$$\lim \cfrac{\|(\cos t,\sin t)-(\cos\xi,\sin\xi)-(-\sin(t-\xi),\cos(t-\xi))\|}{\|t-\xi\|} = $$
$$\lim \cfrac{\|(\cos t - \cos\xi + \sin(t-\xi), \sin t-\sin\xi-\cos(t-\xi))\|}{\|t-\xi\|} = $$
$$\lim\cfrac{\|(\cos t - \cos\xi + \sin t\cos\xi-\cos t\sin\xi, \sin t- \sin\xi - \cos t\cos\xi-\sin t\sin\xi)\|}{\|t-\xi\|}$$
I don't really know where to go from here can anyone help me with this?

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    $\begingroup$ You might simply prove that a function $t\longmapsto \bigl(f(t),g(t)\bigr)$ from $\mathbf R$ to $\mathbf R^2$ is differentiable if and only if each coordinate function $f(t),\,g(t)$ is differentiable. $\endgroup$
    – Bernard
    Sep 23, 2017 at 10:22
  • $\begingroup$ Yes but I also have to show that $A=f'$ so I assumed I had to use this formula, seeing that when the limit is zero it follows that A is the derivative $\endgroup$ Sep 23, 2017 at 10:26
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    $\begingroup$ I'm sorry but $f'=A$ is the definition of $f'$. You don't have to prove it. You have to prove the existence of such an $A$. $\endgroup$
    – Bernard
    Sep 23, 2017 at 10:32

2 Answers 2

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You are misunderstanding what $A(t-\xi)$ means. When you write $A = (-\sin t,\cos t)$, $Ax$ does not mean $(-\sin x,\cos x)$! That function is not even linear. $A = (-\sin t,\cos t)$ is just a matrix representation of linear operator $A$, and in that sense it should actually be $$A=\begin{pmatrix} -\sin t\\ \cos t\end{pmatrix}$$

and $Ax$ should be evaluated by matrix multiplication:

$$Ax = \begin{pmatrix} -\sin t\\ \cos t\end{pmatrix}x = \begin{pmatrix} -x\sin t\\ x\cos t\end{pmatrix},$$ or treating $A$ as linear operator again (and not matrix representation), $Ax = (-x\sin t,x\cos t)$.

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HINT

$f(t)=(f_1(t),f_2(t))$

Now $$\frac{f(t+h)-f(t)}{h}=(\frac{f_1(t+h)-f_1(t)}{h},\frac{f_2(t+h)-f_2(t)}{h})$$

So let $t_0 \in \mathbb{R}, \epsilon>0$ and assume that $\lim_{h \to 0}\frac{f_1(t_0+h)-f_1(t_0)}{h}=y_1=f_1'(t_0)$ and $\lim_{h \to 0}\frac{f_2(t_0+h)-f_2(t_0)}{h}=y_2=f_2'(t_0)$

Thus exists $\delta_1>0$ such that if $|h|< \delta_1$ then $|\frac{f_1(t_0+h)-f_1(t_0)}{h}-f_1'(t_0)|< \frac{\epsilon}{2}$

Also exists $\delta_2>0$ such that if $|h|<\delta_2$ then $|\frac{f_2(t_0+h)-f_2(t_0)}{h}-f_2'(t_0)| < \frac{\epsilon}{2}$

Take $\delta=\min\{\delta_1,\delta_2\}$ and you will have that $$|h|<\delta \Rightarrow ||\frac{f(t_0+h)-f(t_0)}{h}-(y_1,y_2)||_2=\sqrt{(\frac{f_1(t_0+h)-f_1(t_0)}{h}-y_1)^2+(\frac{f_2(t_0+h)-f_2(t_0)}{h}-y_2)^2}< \sqrt{\frac{\epsilon^2}{2}}<\epsilon \Rightarrow \text{f is differentiable at $t_0$}$$

So using this and the fact that $\sin{t},\cos{t}$ are differentiable everywhere,you can prove that $f$ is differentiable.

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