<s>[INST] How to expand $\tan x$ in Taylor order to $o(x^6)$ I try to expand  $\tan x$ in Taylor order to $o(x^6)$, but searching of all 6 derivative in zero (ex. $\tan'(0), \tan''(0)$ and e.t.c.) is very difficult and slow method. 
Is there another way to solve the problem?
Any help would be greatly appreciated :)
 [/INST] This answer is inspired by the answer of coffeemath. 
We know the following (from e.g. Wikipedia):
\begin{align}
&\sin(x) = x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) \\
&\cos(x) = 1 -\frac{x^2}{2} +\frac{x^4}{24} +\frac{x^6}{720} +O(x^8) \\
&\tan(x) = a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)
\end{align}
Furthermore it holds $\tan(x) = \frac{\sin(x)}{\cos(x)} \Rightarrow \sin(x) =\tan(x)\cos(x)$. So we get
\begin{align}
 x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) =& \bigl(1 -\frac{x^2}{2} +\frac{x^4}{24} +\frac{x^6}{720} +O(x^8) \bigr) \\ &\cdot\bigl(a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)\bigr)
\end{align}
If we expand the term on the right hand side we obtain
\begin{align}
 x -\frac{x^3}{6} +\frac{x^5}{120}+ O(x^7) &=a+bx+(-\frac{a}{2}+c)x^2+(-\frac{b}{2}+d)x^3
\\&+(\frac{a}{24}-\frac{c}{2}+e)x^4+(\frac{b}{24}-\frac{d}{2}+f)x^5
\\&+(\frac{a}{720}+\frac{c}{24}-\frac{e}{2}+g)x^6+ O(x^7)
\end{align}
From this we can deduct the linear system
\begin{align}
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0&0\\
0 & 1 & 0 & 0 & 0 & 0 &0\\
-1/2 &0 &1 & 0& 0&  0&0\\
0 & -1/2 & 0 & 1 & 0 & 0&0 \\
1/24 & 0 & -1/2 & 0 & 1 & 0&0\\
0 &1/24 & 0 & -1/2 & 0 & 1&0\\
1/720 & 0 & 1/24 & 0 & -1/2 & 0 &1
\end{pmatrix}
\begin{pmatrix}
a\\b\\c\\d\\e\\f\\g
\end{pmatrix}
=
\begin{pmatrix}
0\\1 \\ 0 \\ -1/6 \\ 0 \\1/120 \\ 0
\end{pmatrix}
\end{align}
Which has the solution
\begin{align}
\begin{pmatrix}
a\\b\\c\\d\\e\\f\\g
\end{pmatrix} 
= \begin{pmatrix}
0 \\ 1 \\ 0 \\ 1/3 \\ 0 \\2/15
\end{pmatrix}
\end{align}
[Note that you can actually extract two linear systems of the one given above. You then only have to solve for the odd powers of $x$ as $\tan(x)$ is odd.]
So you finally can write your Taylor series as:
\begin{align}
&\tan(x) = x+ \frac{1}{3}x^3+\frac{2}{15}x^5+ O(x^7)
\end{align}
Which is (surprisingly) correct, as we can compare our solution with WolframAlpha.
</s>