$n \ge 1$ とする。次の関数 $f(x)$ 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。次の関数 $f(x)$ 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) $f(x) = \frac{3^3}{(1+3x)^4}$

まず、$f(x) = 27(1+3x)^{-4}$ と書き換えます。

$f'(x) = 27(-4)(1+3x)^{-5}(3) = -324(1+3x)^{-5}$

$f''(x) = -324(-5)(1+3x)^{-6}(3) = 4860(1+3x)^{-6}$

$f'''(x) = 4860(-6)(1+3x)^{-7}(3) = -87480(1+3x)^{-7}$

一般に、$f^{(n)}(x) = 27 (-1)^n \frac{(n+3)!}{3!} (3)^n (1+3x)^{-(n+4)}$

したがって、

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$f'''(x) = -64 \sin 4x$

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

$f^{(n)}(x) = (-1)^{\frac{n}{2}} 4^n \cos 4x \quad \text{if n is even}$

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

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

部分分数分解を行う。

$\frac{12}{(x+1)(x+2)(x+3)(x+4)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3} + \frac{D}{x+4}$

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

$x=-1: 12 = A(1)(2)(3) = 6A \Rightarrow A = 2$

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

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

$x=-4: 12 = D(-3)(-2)(-1) = -6D \Rightarrow D = -2$

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

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

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

3. 最終的な答え

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

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

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

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

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