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

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

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

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

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

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

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

$f^{(n)}(x) = 27 \cdot (-1)^n \cdot (n+3)(n+2)(n+1) \cdot 3^n (1+3x)^{-(n+4)}$

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

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

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

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

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

$f(x) = e^{-2x} \cos(2x + \frac{\pi}{18}) = e^{-2x} \cos(2x + \alpha)$とおく。

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

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

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

$4^x = e^{x \ln 4}$

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

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

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

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

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

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

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

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

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

部分分数分解:

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

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

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

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

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

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

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

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

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

3. 最終的な答え

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

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

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

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

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