以下の8つの関数をそれぞれ微分します。 (1) $y = \sin^3(4x)$ (2) $y = \frac{1}{\cos x}$ (3) $y = \sqrt{\tan x}$ (4) $y = e^{-3x}\sin(2x)$ (5) $y = \log |1-x^2|$ (6) $y = \log |\tan x|$ (7) $y = \frac{1}{\log(x^2+1)}$ (8) $y = \frac{\log(1-x^2)}{e^{2x}}$

はい、承知しました。画像にある関数の微分を計算します。

1. 問題の内容

以下の8つの関数をそれぞれ微分します。

(1) $y = \sin^3(4x)$

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

(3) $y = \sqrt{\tan x}$

(4) $y = e^{-3x}\sin(2x)$

(5) $y = \log |1-x^2|$

(6) $y = \log |\tan x|$

(7) $y = \frac{1}{\log(x^2+1)}$

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

2. 解き方の手順

(1) $y = \sin^3(4x)$

合成関数の微分法を用いる。

$y' = 3\sin^2(4x) \cdot (\sin(4x))' = 3\sin^2(4x) \cdot \cos(4x) \cdot (4x)' = 3\sin^2(4x) \cdot \cos(4x) \cdot 4 = 12\sin^2(4x)\cos(4x)$

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

$y' = -1(\cos x)^{-2} \cdot (-\sin x) = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \cdot \sec x$

(3) $y = \sqrt{\tan x} = (\tan x)^{1/2}$

$y' = \frac{1}{2}(\tan x)^{-1/2} \cdot (\tan x)' = \frac{1}{2}(\tan x)^{-1/2} \cdot \frac{1}{\cos^2 x} = \frac{1}{2\sqrt{\tan x}\cos^2 x} = \frac{1}{2\sqrt{\tan x}} (1 + \tan^2 x)$

(4) $y = e^{-3x}\sin(2x)$

積の微分法を用いる。

$y' = (e^{-3x})' \sin(2x) + e^{-3x}(\sin(2x))' = -3e^{-3x}\sin(2x) + e^{-3x}(2\cos(2x)) = e^{-3x}(2\cos(2x) - 3\sin(2x))$

(5) $y = \log |1-x^2|$

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

(6) $y = \log |\tan x|$

$y' = \frac{1}{\tan x} \cdot (\tan x)' = \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{2\sin x \cos x} = \frac{2}{\sin(2x)}$

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

$y' = -1 (\log(x^2+1))^{-2} \cdot (\log(x^2+1))' = -(\log(x^2+1))^{-2} \cdot \frac{1}{x^2+1} \cdot (x^2+1)' = -\frac{1}{(\log(x^2+1))^2} \cdot \frac{2x}{x^2+1} = -\frac{2x}{(x^2+1)(\log(x^2+1))^2}$

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

商の微分法を用いる。

$y' = \frac{(\log(1-x^2))'e^{2x} - \log(1-x^2)(e^{2x})'}{(e^{2x})^2} = \frac{\frac{1}{1-x^2}(-2x)e^{2x} - \log(1-x^2)(2e^{2x})}{e^{4x}} = \frac{e^{2x}(\frac{-2x}{1-x^2} - 2\log(1-x^2))}{e^{4x}} = \frac{\frac{-2x}{1-x^2} - 2\log(1-x^2)}{e^{2x}} = \frac{-2x - 2(1-x^2)\log(1-x^2)}{(1-x^2)e^{2x}}$

3. 最終的な答え

(1) $y' = 12\sin^2(4x)\cos(4x)$

(2) $y' = \tan x \sec x$

(3) $y' = \frac{1}{2\sqrt{\tan x}\cos^2 x} = \frac{1+ \tan^2 x}{2\sqrt{\tan x}}$

(4) $y' = e^{-3x}(2\cos(2x) - 3\sin(2x))$

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

(6) $y' = \frac{2}{\sin(2x)}$

(7) $y' = -\frac{2x}{(x^2+1)(\log(x^2+1))^2}$

(8) $y' = \frac{-2x - 2(1-x^2)\log(1-x^2)}{(1-x^2)e^{2x}}$