次の関数の導関数を求めます。 (7) $y = \tan^{-1}(\sin^{-1}x)$ (8) $y = \log(x + \sqrt{x^2+2})$ (9) $y = \frac{x}{\sqrt{x^2+1}}$ (10) $y = \tan(\frac{x+1}{x^2})$ (11) $y = \sin^3(x^2)$ (12) $y = \tan(\sin(\log x))$

はい、承知いたしました。画像に写っている問題の中から、(7), (8), (9), (10), (11), (12) の関数の導関数を求めます。

1. 問題の内容

次の関数の導関数を求めます。

(7) $y = \tan^{-1}(\sin^{-1}x)$

(8) $y = \log(x + \sqrt{x^2+2})$

(9) $y = \frac{x}{\sqrt{x^2+1}}$

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

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

(12) $y = \tan(\sin(\log x))$

2. 解き方の手順

各問題について、以下の手順で導関数を求めます。

(7) $y = \tan^{-1}(\sin^{-1}x)$

合成関数の微分を行います。

$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \frac{du}{dx}$

$\frac{d}{dx} \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$

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

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

(8) $y = \log(x + \sqrt{x^2+2})$

合成関数の微分を行います。

$\frac{d}{dx} \log(u) = \frac{1}{u} \frac{du}{dx}$

$\frac{d}{dx} \sqrt{x^2+2} = \frac{1}{2\sqrt{x^2+2}} \cdot 2x = \frac{x}{\sqrt{x^2+2}}$

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

$y' = \frac{1}{x + \sqrt{x^2+2}} \cdot \frac{\sqrt{x^2+2} + x}{\sqrt{x^2+2}}$

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

(9) $y = \frac{x}{\sqrt{x^2+1}}$

商の微分を行います。

$(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$

$u = x$, $u' = 1$

$v = \sqrt{x^2+1}$, $v' = \frac{1}{2\sqrt{x^2+1}} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$

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

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

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

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

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

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

合成関数の微分を行います。

$\frac{d}{dx} \tan(u) = \sec^2(u) \frac{du}{dx}$

$(\frac{x+1}{x^2})' = \frac{1 \cdot x^2 - (x+1) \cdot 2x}{(x^2)^2} = \frac{x^2 - 2x^2 - 2x}{x^4} = \frac{-x^2 - 2x}{x^4} = \frac{-x-2}{x^3}$

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

$y' = -\frac{x+2}{x^3} \sec^2(\frac{x+1}{x^2})$

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

合成関数の微分を行います。

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

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

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

(12) $y = \tan(\sin(\log x))$

合成関数の微分を行います。

$\frac{d}{dx} \tan(u) = \sec^2(u) \frac{du}{dx}$

$\frac{d}{dx} \sin(u) = \cos(u) \frac{du}{dx}$

$\frac{d}{dx} \log(x) = \frac{1}{x}$

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

$y' = \frac{\cos(\log x) \sec^2(\sin(\log x))}{x}$

3. 最終的な答え

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

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

(9) $y' = \frac{1}{(x^2+1)^{3/2}}$

(10) $y' = -\frac{x+2}{x^3} \sec^2(\frac{x+1}{x^2})$

(11) $y' = 6x \sin^2(x^2) \cos(x^2)$

(12) $y' = \frac{\cos(\log x) \sec^2(\sin(\log x))}{x}$