次の関数を微分せよ。 (1) $y = \sin^2 3x$ (2) $y = \sin^5 x \cos 5x$ (3) $y = \sin^4 x \cos^4 x$ (4) $y = \sqrt{1+\sin^2 x}$ (5) $y = \sqrt{\sin(x^2+x+1)}$ (6) $y = (\tan x + \frac{1}{\tan x})^2$ (7) $y = \frac{\cos x}{1 - \sin x}$

1. 問題の内容

次の関数を微分せよ。

(1) $y = \sin^2 3x$

(2) $y = \sin^5 x \cos 5x$

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

(4) $y = \sqrt{1+\sin^2 x}$

(5) $y = \sqrt{\sin(x^2+x+1)}$

(6) $y = (\tan x + \frac{1}{\tan x})^2$

(7) $y = \frac{\cos x}{1 - \sin x}$

2. 解き方の手順

(1) $y = \sin^2 3x$ の微分

合成関数の微分法を用いる。

$y' = 2 \sin 3x \cdot (\sin 3x)' = 2 \sin 3x \cdot (\cos 3x) \cdot (3x)' = 2 \sin 3x \cos 3x \cdot 3 = 6 \sin 3x \cos 3x = 3 \sin 6x$

(2) $y = \sin^5 x \cos 5x$ の微分

積の微分法と合成関数の微分法を用いる。

$y' = (\sin^5 x)' \cos 5x + \sin^5 x (\cos 5x)' = 5 \sin^4 x (\sin x)' \cos 5x + \sin^5 x (-\sin 5x) (5x)' = 5 \sin^4 x \cos x \cos 5x - 5 \sin^5 x \sin 5x$

(3) $y = \sin^4 x \cos^4 x = (\sin x \cos x)^4 = (\frac{1}{2} \sin 2x)^4 = \frac{1}{16} \sin^4 2x$ の微分

合成関数の微分法を用いる。

$y' = \frac{1}{16} \cdot 4 \sin^3 2x \cdot (\sin 2x)' = \frac{1}{4} \sin^3 2x \cdot (\cos 2x) \cdot (2x)' = \frac{1}{4} \sin^3 2x \cos 2x \cdot 2 = \frac{1}{2} \sin^3 2x \cos 2x$

(4) $y = \sqrt{1+\sin^2 x} = (1+\sin^2 x)^{\frac{1}{2}}$ の微分

合成関数の微分法を用いる。

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

(5) $y = \sqrt{\sin(x^2+x+1)} = (\sin(x^2+x+1))^{\frac{1}{2}}$ の微分

合成関数の微分法を用いる。

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

(6) $y = (\tan x + \frac{1}{\tan x})^2 = (\tan x + \cot x)^2$ の微分

$y = (\tan x + \cot x)^2 = \tan^2 x + 2 \tan x \cot x + \cot^2 x = \tan^2 x + 2 + \cot^2 x$

$y' = 2 \tan x (\tan x)' + 0 + 2 \cot x (\cot x)' = 2 \tan x \cdot \frac{1}{\cos^2 x} + 2 \cot x \cdot (-\frac{1}{\sin^2 x}) = \frac{2 \sin x}{\cos^3 x} - \frac{2 \cos x}{\sin^3 x} = 2 (\frac{\sin x}{\cos^3 x} - \frac{\cos x}{\sin^3 x}) = 2 \frac{\sin^4 x - \cos^4 x}{\sin^3 x \cos^3 x} = 2 \frac{(\sin^2 x + \cos^2 x)(\sin^2 x - \cos^2 x)}{\sin^3 x \cos^3 x} = 2 \frac{-\cos 2x}{\sin^3 x \cos^3 x} = -2 \frac{\cos 2x}{\sin^3 x \cos^3 x} = -16 \frac{\cos 2x}{\sin^3 2x}$

(7) $y = \frac{\cos x}{1-\sin x}$ の微分

商の微分法を用いる。

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

3. 最終的な答え

(1) $y' = 3 \sin 6x$

(2) $y' = 5 \sin^4 x \cos x \cos 5x - 5 \sin^5 x \sin 5x$

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

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

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

(6) $y' = -16 \frac{\cos 2x}{\sin^3 2x}$

(7) $y' = \frac{1}{1-\sin x}$