How can I prove the even derivatives of $\frac{d^n}{dx^n}\left \{ \right.tan(x)\left. \right \}=0$ at $x=\pi$ How can I prove the even derivatives of $\tan(x)$ 

$$\frac{d^n}{dx^n}\left \{  \right.tan(x)\left.  \right \}=0$$ at $x=\pi$

 A: because the function $\tan(x)$ is odd about $x = \pi$.
added later.
here is the taylor series of $\tan x$ about $x = \pi$ we will use a change of variable $x = pi + u, u = x - pi$
$$\tan x = \tan (\pi + u) = \tan u = u + u^3/3 + 2u^5/15+\cdots = 
(x-\pi) + (x-\pi)^3/3 + 2(x-\pi)^5/15 + \cdots  $$ 
A: For conciseness, let us denote $t=\tan x$.
Then from
$$t'=t^2+1,$$
we have
$$t''=2tt'=2t^3+2t$$
$$t'''=6t't^2+2t'=6t^4+8t^2+1$$
$$...$$
It is an easy matter to generalize that all even/odd derivatives have only odd/even powers.
A: $f(x)=\tan(x)$ is a $\pi$-periodic function, hence $f^{(n)}(\pi)=f^{(n)}(0)$. 
On the other hand, $f(x)$ is an odd function, hence its even derivatives in zero vanish:
$$f^{(2n)}(0)=\frac{1}{2}\frac{d^{2n}}{dx^{2n}}\left.\left(f(x)+f(-x)\right)\right|_{x=0} = \frac{d^{2n}}{dx^{2n}}(0) = 0.$$
A: If you differentiate the relations $f(-x)=f(x)$ and $g(-x)=-g(x)$, you find that the derivative of an even function is an odd function, and the derivative of an odd function is an even function. Therefore, the second derivative of an odd function is odd and hence all even derivatives of an odd function are odd too. 
Since odd functions satisfy $f(0)=0$, if $f$ is odd, $f^{(2n)}(0)=0$. In particular, this holds when $f(x)=\tan(x)$. However, since the tangent function has period $\pi$, so does all its derivatives, and so $\tan^{(2n)}(\pi)=0$
