与えられた8つの関数に対して、$n$次導関数（$n \ge 1$）を求める問題です。

1. 問題の内容

与えられた8つの関数に対して、$n$次導関数（$n \ge 1$）を求める問題です。

2. 解き方の手順

(1) $y = \frac{1}{1+x} = (1+x)^{-1}$

$y' = -1 (1+x)^{-2}$

$y'' = (-1)(-2) (1+x)^{-3}$

$y''' = (-1)(-2)(-3) (1+x)^{-4}$

...

$y^{(n)} = (-1)^n n! (1+x)^{-(n+1)} = \frac{(-1)^n n!}{(1+x)^{n+1}}$

(2) $y = \log(1-x)$

$y' = \frac{-1}{1-x} = -(1-x)^{-1}$

$y'' = -(-1)(-1) (1-x)^{-2} = -(1-x)^{-2}$

$y''' = -(-2) (-1) (1-x)^{-3} = -2 (1-x)^{-3}$

...

$y^{(n)} = -(n-1)! (1-x)^{-n} = -\frac{(n-1)!}{(1-x)^n}$

(3) $y = (1+x)^a$ ($a$は定数)

$y' = a(1+x)^{a-1}$

$y'' = a(a-1)(1+x)^{a-2}$

$y''' = a(a-1)(a-2)(1+x)^{a-3}$

...

$y^{(n)} = a(a-1)(a-2) \dots (a-n+1) (1+x)^{a-n}$

(4) $y = x^2 e^{2x}$

$y' = 2x e^{2x} + x^2 (2e^{2x}) = 2x e^{2x} + 2x^2 e^{2x} = (2x^2 + 2x) e^{2x}$

$y'' = (4x + 2)e^{2x} + (2x^2 + 2x) 2e^{2x} = (4x + 2 + 4x^2 + 4x) e^{2x} = (4x^2 + 8x + 2) e^{2x}$

$y''' = (8x + 8)e^{2x} + (4x^2 + 8x + 2) 2e^{2x} = (8x+8 + 8x^2 + 16x + 4)e^{2x} = (8x^2+24x+12)e^{2x}$

一般のライプニッツの公式を使うと、

$(uv)^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(k)} v^{(n-k)}$

ここで$u = x^2$, $v = e^{2x}$とすると、

$u' = 2x$, $u'' = 2$, $u^{(k)} = 0$ for $k \ge 3$.

$v^{(n)} = 2^n e^{2x}$.

よって、

$y^{(n)} = \binom{n}{0} x^2 2^n e^{2x} + \binom{n}{1} 2x 2^{n-1} e^{2x} + \binom{n}{2} 2 \cdot 2^{n-2} e^{2x} = e^{2x} (x^2 2^n + n \cdot 2x \cdot 2^{n-1} + \frac{n(n-1)}{2} \cdot 2 \cdot 2^{n-2}) = e^{2x} (2^n x^2 + n 2^n x + n(n-1) 2^{n-2}) = 2^{n-2} e^{2x} (4x^2 + 4nx + n(n-1))$

(5) $y = 3x(x^2 + x) = 3x^3 + 3x^2$

$y' = 9x^2 + 6x$

$y'' = 18x + 6$

$y''' = 18$

$y^{(4)} = 0$

...

$y^{(n)} = 0$ for $n \ge 4$.

$y^{(1)} = 9x^2 + 6x$, $y^{(2)} = 18x+6$, $y^{(3)} = 18$.

(6) $y = x^2 \cos(2x)$

$y' = 2x \cos(2x) - 2x^2 \sin(2x)$

$y'' = 2 \cos(2x) - 4x \sin(2x) - 4x \sin(2x) - 4x^2 \cos(2x) = (2 - 4x^2)\cos(2x) - 8x \sin(2x)$

ライプニッツの公式を使う. $u = x^2, v = \cos(2x)$

$u' = 2x$, $u'' = 2$, $u^{(k)} = 0$ for $k \ge 3$.

$v^{(n)} = 2^n \cos(2x + \frac{n\pi}{2})$

$y^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(k)} v^{(n-k)} = \binom{n}{0} x^2 2^n \cos(2x + \frac{n\pi}{2}) + \binom{n}{1} 2x 2^{n-1} \cos(2x + \frac{(n-1)\pi}{2}) + \binom{n}{2} 2 \cdot 2^{n-2} \cos(2x + \frac{(n-2)\pi}{2}) = x^2 2^n \cos(2x + \frac{n\pi}{2}) + n 2^n x \cos(2x + \frac{(n-1)\pi}{2}) + n(n-1) 2^{n-2} \cos(2x + \frac{(n-2)\pi}{2})$

(7) $y = \frac{1}{x^2 - x - 2} = \frac{1}{(x-2)(x+1)} = \frac{1}{3} (\frac{1}{x-2} - \frac{1}{x+1})$

$y' = \frac{1}{3} [-(x-2)^{-2} + (x+1)^{-2}]$

$y'' = \frac{1}{3} [2(x-2)^{-3} - 2(x+1)^{-3}]$

$y''' = \frac{1}{3} [-6(x-2)^{-4} + 6(x+1)^{-4}]$

...

$y^{(n)} = \frac{1}{3} [(-1)^n n! (x-2)^{-(n+1)} - (-1)^n n! (x+1)^{-(n+1)}] = \frac{(-1)^n n!}{3} [ \frac{1}{(x-2)^{n+1}} - \frac{1}{(x+1)^{n+1}}]$

(8) $y = \frac{e^x}{1-x}$

$y' = \frac{e^x (1-x) + e^x}{(1-x)^2} = \frac{e^x - xe^x + e^x}{(1-x)^2} = \frac{2e^x - xe^x}{(1-x)^2}$

ライプニッツの公式を使う. $y= e^x (1-x)^{-1}$

$y^{(n)} = \sum_{k=0}^{n} \binom{n}{k} (e^x)^{(n-k)} ((1-x)^{-1})^{(k)}$

$y^{(n)} = \sum_{k=0}^{n} \binom{n}{k} e^x (-1)^k k! (1-x)^{-(k+1)} = e^x \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k k!}{(1-x)^{k+1}}$

$y^{(n)} = e^x \sum_{k=0}^n \frac{n!}{k! (n-k)!} \frac{(-1)^k k!}{(1-x)^{k+1}} = e^x n! \sum_{k=0}^n \frac{(-1)^k}{(n-k)! (1-x)^{k+1}}$

3. 最終的な答え

(1) $y^{(n)} = \frac{(-1)^n n!}{(1+x)^{n+1}}$

(2) $y^{(n)} = -\frac{(n-1)!}{(1-x)^n}$

(3) $y^{(n)} = a(a-1)(a-2) \dots (a-n+1) (1+x)^{a-n}$

(4) $y^{(n)} = 2^{n-2} e^{2x} (4x^2 + 4nx + n(n-1))$

(5) $y' = 9x^2 + 6x$, $y'' = 18x+6$, $y''' = 18$, $y^{(n)} = 0$ for $n \ge 4$.

(6) $y^{(n)} = x^2 2^n \cos(2x + \frac{n\pi}{2}) + n 2^n x \cos(2x + \frac{(n-1)\pi}{2}) + n(n-1) 2^{n-2} \cos(2x + \frac{(n-2)\pi}{2})$

(7) $y^{(n)} = \frac{(-1)^n n!}{3} [ \frac{1}{(x-2)^{n+1}} - \frac{1}{(x+1)^{n+1}}]$

(8) $y^{(n)} = e^x n! \sum_{k=0}^n \frac{(-1)^k}{(n-k)! (1-x)^{k+1}}$