<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] Here's the long division method suggested by coffemath.
\begin{align}
\sin(x)
= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{(2n+1)}
&=& \sum_{n=0}^\infty a_{(2n+1)} x^{(2n+1)} \\
\cos(x)
= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
&= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{(2n)}
&=& \sum_{n=0}^\infty b_{(2n)} x^{(2n)} \\
\end{align}

\begin{align}
\tan(x) = \frac{\sin(x)}{\cos(x)}
\end{align}

\begin{align}
\sin(x) - \tan(x)\cos(x) &= 0 \\
\sin(x) - truncated(\tan(x))\cos(x) &= remainder\\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) - 
\left( 0 \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) - 
\left( x \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{x^3}{3} - \frac{x^5}{30} + \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) - 
\left( x + \frac{x^3}{3} \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{2x^5}{15} - \frac{4x^7}{315} + \dots \right) \\
\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \right) - 
\left( x + \frac{x^3}{3} + \frac{2x^5}{15} \right)
\left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \right)
&= \left( \frac{17x^7}{315} - \frac{29x^9}{5670} + \dots \right)
\end{align}

\begin{align}
\tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \dots = \sum_{n=0}^{\infty}c_{(2n+1)}x^{(2n+1)}
\end{align}
Where the coefficients are given recursively as:
\begin{align}
a_{(2n+1)} &= \frac{(-1)^n}{(2n+1)!} \\
b_{(2n)} &= \frac{(-1)^n}{(2n)!} \\
c_{(2n+1)} &= a_{(2n+1)} - \sum_{k=0}^{n-1} b_{(2n-2k)} c_{(2k+1)}. \quad \blacksquare
\end{align}
This series converges with radius $\frac{\pi}{2}$ around $x=0$.
</s>