与えられた6つの関数それぞれについて、第3次導関数までを求める問題です。 (1) $y = x^3 - 2x + 5$ (2) $y = \frac{1}{x}$ (3) $y = \cos x$ (4) $y = \log x$ (5) $y = e^x$ (6) $y = e^{-2x}$

1. 問題の内容

与えられた6つの関数それぞれについて、第3次導関数までを求める問題です。

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

(2) $y = \frac{1}{x}$

(3) $y = \cos x$

(4) $y = \log x$

(5) $y = e^x$

(6) $y = e^{-2x}$

2. 解き方の手順

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

- 1次導関数: $y' = 3x^2 - 2$

- 2次導関数: $y'' = 6x$

- 3次導関数: $y''' = 6$

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

- 1次導関数: $y' = -1x^{-2} = -\frac{1}{x^2}$

- 2次導関数: $y'' = 2x^{-3} = \frac{2}{x^3}$

- 3次導関数: $y''' = -6x^{-4} = -\frac{6}{x^4}$

(3) $y = \cos x$

- 1次導関数: $y' = -\sin x$

- 2次導関数: $y'' = -\cos x$

- 3次導関数: $y''' = \sin x$

(4) $y = \log x$

- 1次導関数: $y' = \frac{1}{x}$

- 2次導関数: $y'' = -\frac{1}{x^2}$

- 3次導関数: $y''' = \frac{2}{x^3}$

(5) $y = e^x$

- 1次導関数: $y' = e^x$

- 2次導関数: $y'' = e^x$

- 3次導関数: $y''' = e^x$

(6) $y = e^{-2x}$

- 1次導関数: $y' = -2e^{-2x}$

- 2次導関数: $y'' = 4e^{-2x}$

- 3次導関数: $y''' = -8e^{-2x}$

3. 最終的な答え

(1)

$y' = 3x^2 - 2$

$y'' = 6x$

$y''' = 6$

(2)

$y' = -\frac{1}{x^2}$

$y'' = \frac{2}{x^3}$

$y''' = -\frac{6}{x^4}$

(3)

$y' = -\sin x$

$y'' = -\cos x$

$y''' = \sin x$

(4)

$y' = \frac{1}{x}$

$y'' = -\frac{1}{x^2}$

$y''' = \frac{2}{x^3}$

(5)

$y' = e^x$

$y'' = e^x$

$y''' = e^x$

(6)

$y' = -2e^{-2x}$

$y'' = 4e^{-2x}$

$y''' = -8e^{-2x}$