逆三角関数 $arccos(\frac{1+2sin(x)}{2+sin(x)})$ を微分せよ。

1. 問題の内容

逆三角関数 $arccos(\frac{1+2sin(x)}{2+sin(x)})$ を微分せよ。

2. 解き方の手順

関数 $y = arccos(u)$ の微分は $\frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}$ であり、

$u = \frac{1+2sin(x)}{2+sin(x)}$ とおく。

まず、$u$ の微分を計算する。

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

$\frac{du}{dx} = \frac{4cos(x) + 2sin(x)cos(x) - cos(x) - 2sin(x)cos(x)}{(2+sin(x))^2}$

$\frac{du}{dx} = \frac{3cos(x)}{(2+sin(x))^2}$

次に、$1-u^2$ を計算する。

$1-u^2 = 1 - (\frac{1+2sin(x)}{2+sin(x)})^2 = \frac{(2+sin(x))^2 - (1+2sin(x))^2}{(2+sin(x))^2}$

$1-u^2 = \frac{4+4sin(x)+sin^2(x) - (1+4sin(x)+4sin^2(x))}{(2+sin(x))^2}$

$1-u^2 = \frac{3 - 3sin^2(x)}{(2+sin(x))^2} = \frac{3(1-sin^2(x))}{(2+sin(x))^2} = \frac{3cos^2(x)}{(2+sin(x))^2}$

したがって、 $\sqrt{1-u^2} = \frac{\sqrt{3}|cos(x)|}{2+sin(x)}$

よって、

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \frac{du}{dx} = -\frac{2+sin(x)}{\sqrt{3}|cos(x)|} \cdot \frac{3cos(x)}{(2+sin(x))^2}$

$\frac{dy}{dx} = -\frac{3cos(x)}{\sqrt{3}|cos(x)|(2+sin(x))}$

$\frac{dy}{dx} = -\frac{\sqrt{3}sign(cos(x))}{2+sin(x)}$

3. 最終的な答え

$-\frac{\sqrt{3}sign(cos(x))}{2+sin(x)}$