与えられた6つの関数について、それぞれ$n$次導関数を求める問題です。 (1) $f(x) = \frac{x+a}{x+b}$ (2) $f(x) = \frac{1}{x^2 - 3x + 2}$ (3) $f(x) = \frac{x^4}{1-x}$ (4) $f(x) = \sin^3 x$ (5) $f(x) = \sin x \cos^3 x$ (6) $f(x) = x^2 \sin x$

1. 問題の内容

与えられた6つの関数について、それぞれ$n$次導関数を求める問題です。

(1) $f(x) = \frac{x+a}{x+b}$

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

(3) $f(x) = \frac{x^4}{1-x}$

(4) $f(x) = \sin^3 x$

(5) $f(x) = \sin x \cos^3 x$

(6) $f(x) = x^2 \sin x$

2. 解き方の手順

(1) $f(x) = \frac{x+a}{x+b} = \frac{x+b+a-b}{x+b} = 1 + \frac{a-b}{x+b}$

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

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

$f'''(x) = (a-b)(-1)(-2)(-3)(x+b)^{-4}$

一般的に、

$f^{(n)}(x) = (a-b)(-1)^n n! (x+b)^{-(n+1)} = (a-b)\frac{(-1)^n n!}{(x+b)^{n+1}}$

(2) $f(x) = \frac{1}{x^2 - 3x + 2} = \frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$

$1 = A(x-2) + B(x-1)$

$x=1$のとき、 $1 = A(1-2) \Rightarrow A = -1$

$x=2$のとき、 $1 = B(2-1) \Rightarrow B = 1$

$f(x) = -\frac{1}{x-1} + \frac{1}{x-2} = -(x-1)^{-1} + (x-2)^{-1}$

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

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

$f^{(n)}(x) = -(-1)^n n! (x-1)^{-(n+1)} + (-1)^n n! (x-2)^{-(n+1)} = (-1)^n n! \left(-\frac{1}{(x-1)^{n+1}} + \frac{1}{(x-2)^{n+1}}\right)$

(3) $f(x) = \frac{x^4}{1-x}$

多項式除算を行うと

$x^4 = -(x^3 + x^2 + x + 1)(1-x) + 1$

$\frac{x^4}{1-x} = -x^3 - x^2 - x - 1 + \frac{1}{1-x}$

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

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

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

$f^{(n)}(x) = -0 - 0 - 0 + (-1)^n n! (1-x)^{-(n+1)} = \frac{n!}{(1-x)^{n+1}} , n\ge4$

$f^{(n)}(x) = \frac{d^n}{dx^n}(-x^3 - x^2 - x - 1) + \frac{d^n}{dx^n}(\frac{1}{1-x})$

$n < 4$の時、

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

$f^{(2)}(x) = -6x-2 + \frac{2}{(1-x)^3}$

$f^{(3)}(x) = -6 + \frac{6}{(1-x)^4}$

$n \ge 4$の時、

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

(4) $f(x) = \sin^3 x = \sin x \sin^2 x = \sin x (\frac{1 - \cos 2x}{2}) = \frac{1}{2}\sin x - \frac{1}{2}\sin x \cos 2x = \frac{1}{2}\sin x - \frac{1}{2} \frac{1}{2} (\sin 3x - \sin x) = \frac{1}{2}\sin x - \frac{1}{4}\sin 3x + \frac{1}{4}\sin x = \frac{3}{4}\sin x - \frac{1}{4}\sin 3x$

$f^{(n)}(x) = \frac{3}{4}\sin(x + \frac{n\pi}{2}) - \frac{1}{4}3^n\sin(3x + \frac{n\pi}{2})$

(5) $f(x) = \sin x \cos^3 x = \sin x \cos x \cos^2 x = \frac{1}{2}\sin 2x \frac{1 + \cos 2x}{2} = \frac{1}{4}\sin 2x + \frac{1}{4}\sin 2x \cos 2x = \frac{1}{4}\sin 2x + \frac{1}{4}\frac{1}{2}\sin 4x = \frac{1}{4}\sin 2x + \frac{1}{8}\sin 4x$

$f^{(n)}(x) = \frac{1}{4}2^n \sin(2x + \frac{n\pi}{2}) + \frac{1}{8}4^n \sin(4x + \frac{n\pi}{2}) = 2^{n-2} \sin(2x + \frac{n\pi}{2}) + 2^{2n-3} \sin(4x + \frac{n\pi}{2})$

(6) $f(x) = x^2 \sin x$

$f'(x) = 2x \sin x + x^2 \cos x$

$f''(x) = 2 \sin x + 2x \cos x + 2x \cos x - x^2 \sin x = 2 \sin x + 4x \cos x - x^2 \sin x$

$f'''(x) = 2 \cos x + 4 \cos x - 4x \sin x - 2x \sin x - x^2 \cos x = 6 \cos x - 6x \sin x - x^2 \cos x$

$f^{(4)}(x) = -6 \sin x - 6 \sin x - 6x \cos x + x^2 \sin x - 2x \cos x = -12 \sin x - 8x \cos x + x^2 \sin x$

ライプニッツの公式を用いる。

$f(x) = x^2 \sin x$

$f^{(n)}(x) = \sum_{k=0}^{n} \binom{n}{k} (x^2)^{(k)} (\sin x)^{(n-k)}$

$= \binom{n}{0}x^2 (\sin x)^{(n)} + \binom{n}{1}2x (\sin x)^{(n-1)} + \binom{n}{2}2 (\sin x)^{(n-2)} + 0$

$= x^2 \sin(x + \frac{n\pi}{2}) + 2nx \sin(x + \frac{(n-1)\pi}{2}) + n(n-1) \sin(x + \frac{(n-2)\pi}{2})$

3. 最終的な答え

(1) $f^{(n)}(x) = (a-b)\frac{(-1)^n n!}{(x+b)^{n+1}}$

(2) $f^{(n)}(x) = (-1)^n n! \left(-\frac{1}{(x-1)^{n+1}} + \frac{1}{(x-2)^{n+1}}\right)$

(3) $f^{(n)}(x) = \frac{n!}{(1-x)^{n+1}} , n\ge4$

(4) $f^{(n)}(x) = \frac{3}{4}\sin(x + \frac{n\pi}{2}) - \frac{1}{4}3^n\sin(3x + \frac{n\pi}{2})$

(5) $f^{(n)}(x) = 2^{n-2} \sin(2x + \frac{n\pi}{2}) + 2^{2n-3} \sin(4x + \frac{n\pi}{2})$

(6) $f^{(n)}(x) = x^2 \sin(x + \frac{n\pi}{2}) + 2nx \sin(x + \frac{(n-1)\pi}{2}) + n(n-1) \sin(x + \frac{(n-2)\pi}{2})$