easy极限

1,y＝arctanx，求yn(0)； 注：yn表示y的n阶导数；

n = 0, y^(0)(x) = arctanx, y^(0)(0) = 0.
n = 1, y'(x) = 1/[1+x^2],
y'(0) = 1.
又
[f(x)g(x)]^(n) =
f^(n)(x)g(x) + nf^(n-1)(x)g^(1)(x) +
+ n(n-1)/2 f^(n-2)(x)g^(2)(x) + ... +
+ nf^(1)(x)g^(n-1)(x) + f(x)g^(n)(x)

而 [1+x^2](n) = 1 + x^2, n = 0.
= 2x , n = 1.
= 2 , n = 2.
= 0 , n > 2.

由 y'(x) = 1/[1+x^2],有
y'(x)[1 + x^2] = 1
上式等号两边分别求n阶导数(n>=1)，有
y^(n+1)(x)[1 + x^2] + 2nxy^(n)(x) + n(n-1)y^(n-1)(x) = 0

令x = 0,有
y^(n+1)(0) + n(n-1)y^(n-1)(0) = 0

y^(n+1)(0) = -n(n-1)y^(n-1)(0)

令n=1,有
y^(2)(0) = 0

由归纳法可以证明，

y^(2n)(0) = 0. 当n为正整数时恒成立。

令n=2,由y^(n+1)(0) = -n(n-1)y^(n-1)(0)
，有

y^(3)(0) = -2y^(1)(0) = -2 = -2!

y^(5)(0) = -4*3y^(3)(0)
= -4*3*(-2)
= 4!
...
由归纳法可以证明，

y^(2n+1)(0) = (-1)