はい、承知いたしました。問題文を読み解き、各問題の解き方と答えをまとめます。

はい、承知いたしました。問題文を読み解き、各問題の解き方と答えをまとめます。

1. 問題の内容

3. 逆三角関数の値を求める問題

(1) $\sin^{-1}(\frac{1}{2})$

(2) $\sin^{-1}(-\frac{1}{2})$

(3) $\sin^{-1}(1)$

(4) $\cos^{-1}(0)$

(5) $\cos^{-1}(1)$

(6) $\cos^{-1}(\frac{1}{\sqrt{2}})$

(7) $\tan^{-1}(\frac{1}{\sqrt{3}})$

(8) $\tan^{-1}(-1)$

(9) $\tan^{-1}(\infty)$

4. 関数を微分する問題

(1) $\cos(4x)$

(2) $x\sin x$

(3) $\sin x\cos x$

(4) $\cos(\sin x)$

(5) $\frac{1}{\sin x}$

(6) $\frac{1}{\tan x}$

5. 関数を微分する問題

(1) $\tan^{-1}(\frac{x-1}{x+1})$

(2) $\sin^{-1}(e^{-x^2})$

(3) $\tan^{-1}(e^x+e^{-x})$

6. 極限を求める問題

(1) $\lim_{x\to 0} \frac{\sin x}{\sin 2x}$

(2) $\lim_{x\to 0} \frac{\tan x}{x}$

(3) $\lim_{x\to 0} \frac{x}{\sin^{-1} x}$

7. 解き方の手順と答え

8. 逆三角関数の値を求める

(1) $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$

$\sin(\frac{\pi}{6})=\frac{1}{2}$

(2) $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$

$\sin(-\frac{\pi}{6})=-\frac{1}{2}$

(3) $\sin^{-1}(1) = \frac{\pi}{2}$

$\sin(\frac{\pi}{2})=1$

(4) $\cos^{-1}(0) = \frac{\pi}{2}$

$\cos(\frac{\pi}{2})=0$

(5) $\cos^{-1}(1) = 0$

$\cos(0)=1$

(6) $\cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$

$\cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$

(7) $\tan^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$

$\tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}$

(8) $\tan^{-1}(-1) = -\frac{\pi}{4}$

$\tan(-\frac{\pi}{4})=-1$

(9) $\tan^{-1}(\infty) = \frac{\pi}{2}$

$\lim_{x \to \frac{\pi}{2}} \tan(x)=\infty$

9. 関数の微分

(1) $(\cos(4x))' = -4\sin(4x)$

(2) $(x\sin x)' = \sin x + x\cos x$

(3) $(\sin x\cos x)' = \cos^2 x - \sin^2 x = \cos 2x$

(4) $(\cos(\sin x))' = -\sin(\sin x) \cdot \cos x$

(5) $(\frac{1}{\sin x})' = (\csc x)' = -\csc x \cot x = -\frac{\cos x}{\sin^2 x}$

(6) $(\frac{1}{\tan x})' = (\cot x)' = -\csc^2 x = -\frac{1}{\sin^2 x}$

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0. 関数の微分

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

(2) $(\sin^{-1}(e^{-x^2}))' = \frac{1}{\sqrt{1-(e^{-x^2})^2}} \cdot (-2xe^{-x^2}) = \frac{-2xe^{-x^2}}{\sqrt{1-e^{-2x^2}}}$

(3) $(\tan^{-1}(e^x+e^{-x}))' = \frac{1}{1+(e^x+e^{-x})^2} \cdot (e^x-e^{-x}) = \frac{e^x-e^{-x}}{1+(e^x+e^{-x})^2}$

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1. 極限

(1) $\lim_{x\to 0} \frac{\sin x}{\sin 2x} = \lim_{x\to 0} \frac{\sin x}{2\sin x \cos x} = \lim_{x\to 0} \frac{1}{2\cos x} = \frac{1}{2\cos 0} = \frac{1}{2}$

(2) $\lim_{x\to 0} \frac{\tan x}{x} = \lim_{x\to 0} \frac{\sin x}{x\cos x} = \lim_{x\to 0} \frac{\sin x}{x} \cdot \lim_{x\to 0} \frac{1}{\cos x} = 1 \cdot 1 = 1$

(3) $\lim_{x\to 0} \frac{x}{\sin^{-1} x} = 1$ ( $y=\sin^{-1} x$とおくと、$x=\sin y$。$x \to 0$のとき、$y \to 0$。 よって$\lim_{x\to 0} \frac{x}{\sin^{-1} x}=\lim_{y\to 0} \frac{\sin y}{y} = 1$)

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2. 最終的な答え

3. (1) $ \frac{\pi}{6} $

(2) $-\frac{\pi}{6}$

(3) $\frac{\pi}{2}$

(4) $\frac{\pi}{2}$

(5) $0$

(6) $\frac{\pi}{4}$

(7) $\frac{\pi}{6}$

(8) $-\frac{\pi}{4}$

(9) $\frac{\pi}{2}$

4. (1) $ -4\sin(4x) $

(2) $\sin x + x\cos x$

(3) $\cos 2x$

(4) $-\sin(\sin x) \cdot \cos x$

(5) $-\frac{\cos x}{\sin^2 x}$

(6) $-\frac{1}{\sin^2 x}$

5. (1) $ \frac{1}{x^2+1} $

(2) $\frac{-2xe^{-x^2}}{\sqrt{1-e^{-2x^2}}}$

(3) $\frac{e^x-e^{-x}}{1+(e^x+e^{-x})^2}$

6. (1) $ \frac{1}{2} $

(2) $1$

(3) $1$