$n \ge 1$ とする。以下の関数 $f(x)$ の $n$ 階導関数 $f^{(n)}(x)$ を求めよ。 1) $f(x) = \frac{3^3}{(1+3x)^4}$ 2) $f(x) = e^{-2x} (2 \cos^2(x+\frac{\pi}{36}) - 1)$ 3) $f(x) = x^2 \cdot 4^x$ 4) $f(x) = 8 \sin^2 x (1 - \sin^2 x)$ 5) $f(x) = \frac{12}{(x+1)(x+2)(x+3)(x+4)}$

1. 問題の内容

$n \ge 1$ とする。以下の関数 $f(x)$ の $n$ 階導関数 $f^{(n)}(x)$ を求めよ。

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

2) $f(x) = e^{-2x} (2 \cos^2(x+\frac{\pi}{36}) - 1)$

3) $f(x) = x^2 \cdot 4^x$

4) $f(x) = 8 \sin^2 x (1 - \sin^2 x)$

5) $f(x) = \frac{12}{(x+1)(x+2)(x+3)(x+4)}$

2. 解き方の手順

1) $f(x) = \frac{3^3}{(1+3x)^4} = 27 (1+3x)^{-4}$

$f'(x) = 27(-4)(1+3x)^{-5}(3) = 27(-4)(3)(1+3x)^{-5}$

$f''(x) = 27(-4)(-5)(3)^2 (1+3x)^{-6}$

$f^{(n)}(x) = 27(-1)^n (4)(5) \cdots (n+3) (3)^n (1+3x)^{-(n+4)} = 27(-1)^n \frac{(n+3)!}{3!} 3^n (1+3x)^{-(n+4)}$

$f^{(n)}(x) = 27(-1)^n \frac{(n+3)!}{6} 3^n (1+3x)^{-(n+4)} = (-1)^n \frac{9}{2} (n+3)! 3^n (1+3x)^{-(n+4)}$

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

2) $f(x) = e^{-2x} (2 \cos^2(x+\frac{\pi}{36}) - 1) = e^{-2x} \cos(2x + \frac{\pi}{18})$

$f'(x) = -2 e^{-2x} \cos(2x+\frac{\pi}{18}) - 2 e^{-2x} \sin(2x+\frac{\pi}{18}) = 2 e^{-2x} (-\cos(2x+\frac{\pi}{18}) - \sin(2x+\frac{\pi}{18}))$

$f'(x) = 2 e^{-2x} (-\cos(2x+\frac{\pi}{18}) - \sin(2x+\frac{\pi}{18})) = 2 e^{-2x} \sqrt{2} (-\frac{1}{\sqrt{2}}\cos(2x+\frac{\pi}{18}) - \frac{1}{\sqrt{2}} \sin(2x+\frac{\pi}{18})) = 2\sqrt{2} e^{-2x} \cos(2x+\frac{\pi}{18} + \frac{3\pi}{4})$

$f'(x) = 2\sqrt{2} e^{-2x} \cos(2x+\frac{\pi}{18} + \frac{5\pi}{4})$

$f^{(n)}(x) = (2\sqrt{2})^n e^{-2x} \cos(2x + \frac{\pi}{18} + \frac{n \pi}{4}) = 2^{n/2} e^{-2x} 2^{n/2} \cos(2x+\frac{\pi}{18}+\frac{n\pi}{4})$

$f^{(n)}(x) = 2^{n} e^{-2x} \cos(2x+\frac{\pi}{18}+\frac{n\pi}{4})$

3) $f(x) = x^2 \cdot 4^x = x^2 e^{x \ln 4}$

$f'(x) = 2x \cdot 4^x + x^2 \cdot 4^x \ln 4 = 4^x (x^2 \ln 4 + 2x)$

$f''(x) = 4^x \ln 4 (x^2 \ln 4 + 2x) + 4^x (2x \ln 4 + 2)$

$f''(x) = 4^x (x^2 (\ln 4)^2 + 4x \ln 4 + 2)$

$f'''(x) = 4^x \ln 4 (x^2 (\ln 4)^2 + 4x \ln 4 + 2) + 4^x (2x (\ln 4)^2 + 4 \ln 4)$

$f'''(x) = 4^x (x^2 (\ln 4)^3 + 6 x (\ln 4)^2 + 6 \ln 4)$

$f^{(n)}(x) = 4^x (x^2 (\ln 4)^n + 2nx (\ln 4)^{n-1} + n(n-1) (\ln 4)^{n-2})$

4) $f(x) = 8 \sin^2 x (1 - \sin^2 x) = 8 \sin^2 x \cos^2 x = 2 (2 \sin x \cos x)^2 = 2 (\sin 2x)^2 = 2 \cdot \frac{1 - \cos 4x}{2} = 1 - \cos 4x$

$f'(x) = 4 \sin 4x$

$f''(x) = 16 \cos 4x$

$f^{(n)}(x) = 4^n (-\cos (4x + n \frac{\pi}{2}))$ if n is even.

$f^{(n)}(x) = 4^n \sin (4x + (n-1) \frac{\pi}{2}))$ if n is odd.

$f^{(n)}(x) = 4^n (-\cos 4x) = -4^n \cos 4x$ when n is even, and

$f^{(n)}(x) = 4^n \sin 4x$ when n is odd.

Therefore, $f^{(n)}(x) = 4^n \sin (4x + \frac{n\pi}{2})$

5) $f(x) = \frac{12}{(x+1)(x+2)(x+3)(x+4)}$

$f(x) = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3} + \frac{D}{x+4}$

$12 = A(x+2)(x+3)(x+4) + B(x+1)(x+3)(x+4) + C(x+1)(x+2)(x+4) + D(x+1)(x+2)(x+3)$

$x = -1: 12 = A(1)(2)(3) \implies A = 2$

$x = -2: 12 = B(-1)(1)(2) \implies B = -6$

$x = -3: 12 = C(-2)(-1)(1) \implies C = 6$

$x = -4: 12 = D(-3)(-2)(-1) \implies D = -2$

$f(x) = \frac{2}{x+1} - \frac{6}{x+2} + \frac{6}{x+3} - \frac{2}{x+4}$

$f^{(n)}(x) = 2 (-1)^n n! (x+1)^{-(n+1)} - 6(-1)^n n! (x+2)^{-(n+1)} + 6(-1)^n n! (x+3)^{-(n+1)} - 2(-1)^n n! (x+4)^{-(n+1)}$

$f^{(n)}(x) = (-1)^n n! \left[ \frac{2}{(x+1)^{n+1}} - \frac{6}{(x+2)^{n+1}} + \frac{6}{(x+3)^{n+1}} - \frac{2}{(x+4)^{n+1}} \right]$

3. 最終的な答え

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

2) $f^{(n)}(x) = 2^{n} e^{-2x} \cos(2x+\frac{\pi}{18}+\frac{n\pi}{4})$

3) $f^{(n)}(x) = 4^x (x^2 (\ln 4)^n + 2nx (\ln 4)^{n-1} + n(n-1) (\ln 4)^{n-2})$

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

5) $f^{(n)}(x) = (-1)^n n! \left[ \frac{2}{(x+1)^{n+1}} - \frac{6}{(x+2)^{n+1}} + \frac{6}{(x+3)^{n+1}} - \frac{2}{(x+4)^{n+1}} \right]$