TITLE: Derivative $o(x)$
QUESTION [0 upvotes]: I'm not sure what I'm writing about, please correct me if I'm wrong.
These are my hypothesis:
\begin{align*}
\cos(x)&=1+o(x) ~~ x \to 0 \\
\cos'(x)&=-\sin(x) \\
\cos''(x)&=-\cos(x)
\end{align*}
If I take the derivative of $1+o(x)$, the result should be an approximation of $-\sin(0)$, right?
If I consider $[1+o(x)]'$ equals to $o(x)$ or $o(1)$ it doesn't make any difference beacuse $\sin(0)=0=o(1)=o(x)$.  
If I reiterate the same process to get the second derivative, the problem gets more interesting.
The second derivative of $1+o(x)$ should be an approximation of $-\cos(0)$.
If I consider 
\begin{align*}
o'(x)=o(x) &\implies \cos'(0)=[1+o(x)]'=o(x)=-\sin(0) \\
&\implies \cos''(0)=[cos'(0)]'=o'(x)=o(x)=-\cos(0) \Rightarrow\!\Leftarrow \\
\end{align*}
So my guess is that $o'(x)=o(1)$
\begin{align*}
o'(x)=o(1) &\implies \cos'(0)=[1+o(x)]'=o(1)=-\sin(0) \\
&\implies \cos''(0)=[cos'(0)]'=o'(1)=[everything]=-\cos(0) 
\end{align*}
P.S: 
\begin{align*}
o'(f(x))&=\{g'(x)|g(x)=o(f(x))\} \\
o(1)&=\{f(x)|\lim_{x \to x_{0}} f(x) = 0\}
\end{align*}
These two equations implies that $o'(1)=[everything]$.
I really appreciate any kind of help.

REPLY [0 votes]: In general, you can't differentiate little $o$s. Consider, for instance, $$x^2\sin\left(\frac1{x^2}\right)=o(x)$$ and its derivative. However, if you know that $f$ is increasing, then this is allowable -- and $\sin$ is monotonic near $0$, so this is fine.
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