次の関数の微分を求めます。 (1) $y = \log_a x \quad (a > 0)$ (2) $y = \sin(\frac{1}{x})$ (3) $y = (\arcsin x)(\arccos x)$ (4) $y = \arctan(\frac{x+2}{3})$ (5) $y = (\frac{x-2}{x^2+1})^3$ (6) $y = \sqrt{\frac{1+\sin x}{1-\sin x}}$ (7) $y = \log(x + \sqrt{x^2+1})$ (8) $y = \frac{1}{2}\{x\sqrt{x^2+1} + \log(x + \sqrt{x^2+1})\}$ (9) $y = \sqrt{\frac{1-x}{1+x}}$ (10) $y = \arcsin(\frac{1}{x})$ (11) $y = 2^x$ (12) $y = e^{x^2}$

はい、承知いたしました。与えられた数学の問題を解いていきます。

1. 問題の内容

次の関数の微分を求めます。

(1) $y = \log_a x \quad (a > 0)$

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

(3) $y = (\arcsin x)(\arccos x)$

(4) $y = \arctan(\frac{x+2}{3})$

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

(6) $y = \sqrt{\frac{1+\sin x}{1-\sin x}}$

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

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

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

(10) $y = \arcsin(\frac{1}{x})$

(11) $y = 2^x$

(12) $y = e^{x^2}$

2. 解き方の手順

(1) $y = \log_a x$

$\frac{dy}{dx} = \frac{1}{x \ln a}$

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

$\frac{dy}{dx} = \cos(\frac{1}{x}) \cdot (-\frac{1}{x^2}) = -\frac{1}{x^2}\cos(\frac{1}{x})$

(3) $y = (\arcsin x)(\arccos x)$

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \arccos x + \arcsin x (-\frac{1}{\sqrt{1-x^2}}) = \frac{\arccos x - \arcsin x}{\sqrt{1-x^2}}$

(4) $y = \arctan(\frac{x+2}{3})$

$\frac{dy}{dx} = \frac{1}{1+(\frac{x+2}{3})^2} \cdot \frac{1}{3} = \frac{1}{3} \cdot \frac{1}{1+\frac{(x+2)^2}{9}} = \frac{1}{3} \cdot \frac{9}{9+(x+2)^2} = \frac{3}{9+(x+2)^2} = \frac{3}{9+x^2+4x+4} = \frac{3}{x^2+4x+13}$

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

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

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

場合分けが必要ですが、ここでは $\cos x > 0$ として計算を進めます。

$y = \frac{1+\sin x}{\cos x} = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \sec x + \tan x$

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

$\cos x < 0$ のときは $y=-\frac{1+\sin x}{\cos x}$ となるため， $\frac{dy}{dx}=-\frac{1}{1-\sin x}$. よって, $\frac{dy}{dx} = \frac{1}{1-\sin x}$ if $\cos x >0$ and $\frac{dy}{dx} = -\frac{1}{1-\sin x}$ if $\cos x < 0$.

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

$\frac{dy}{dx} = \frac{1}{x+\sqrt{x^2+1}} \cdot (1 + \frac{1}{2}(x^2+1)^{-1/2}(2x)) = \frac{1}{x+\sqrt{x^2+1}} \cdot (1 + \frac{x}{\sqrt{x^2+1}}) = \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}$

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

$\frac{dy}{dx} = \frac{1}{2}\{\sqrt{x^2+1} + x \cdot \frac{1}{2}(x^2+1)^{-1/2}(2x) + \frac{1}{x+\sqrt{x^2+1}}(1+\frac{x}{\sqrt{x^2+1}})\} = \frac{1}{2}\{\sqrt{x^2+1} + \frac{x^2}{\sqrt{x^2+1}} + \frac{1}{\sqrt{x^2+1}}\} = \frac{1}{2}\{\frac{x^2+1+x^2+1}{\sqrt{x^2+1}}\} = \frac{2x^2+2}{2\sqrt{x^2+1}} = \frac{x^2+1}{\sqrt{x^2+1}} = \sqrt{x^2+1}$

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

$\frac{dy}{dx} = \frac{1}{2}(\frac{1-x}{1+x})^{-1/2} \cdot \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{1}{2}\sqrt{\frac{1+x}{1-x}} \cdot \frac{-1-x-1+x}{(1+x)^2} = \frac{1}{2}\sqrt{\frac{1+x}{1-x}} \cdot \frac{-2}{(1+x)^2} = -\sqrt{\frac{1+x}{1-x}} \cdot \frac{1}{(1+x)^2} = -\frac{1}{\sqrt{(1-x)(1+x)^3}} = -\frac{1}{(1+x)\sqrt{1-x^2}}$

(10) $y = \arcsin(\frac{1}{x})$

$\frac{dy}{dx} = \frac{1}{\sqrt{1-(\frac{1}{x})^2}} \cdot (-\frac{1}{x^2}) = \frac{1}{\sqrt{1-\frac{1}{x^2}}} \cdot (-\frac{1}{x^2}) = \frac{1}{\sqrt{\frac{x^2-1}{x^2}}} \cdot (-\frac{1}{x^2}) = \frac{|x|}{\sqrt{x^2-1}} \cdot (-\frac{1}{x^2}) = -\frac{|x|}{x^2\sqrt{x^2-1}}$

(11) $y = 2^x$

$\frac{dy}{dx} = 2^x \ln 2$

(12) $y = e^{x^2}$

$\frac{dy}{dx} = e^{x^2} \cdot 2x = 2xe^{x^2}$

3. 最終的な答え

(1) $\frac{dy}{dx} = \frac{1}{x \ln a}$

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

(3) $\frac{dy}{dx} = \frac{\arccos x - \arcsin x}{\sqrt{1-x^2}}$

(4) $\frac{dy}{dx} = \frac{3}{x^2+4x+13}$

(5) $\frac{dy}{dx} = \frac{3(x-2)^2(-x^2+4x+1)}{(x^2+1)^4}$

(6) $\frac{dy}{dx} = \frac{1}{1-\sin x}$, if $\cos x > 0$, otherwise $\frac{dy}{dx} = -\frac{1}{1-\sin x}$

(7) $\frac{dy}{dx} = \frac{1}{\sqrt{x^2+1}}$

(8) $\frac{dy}{dx} = \sqrt{x^2+1}$

(9) $\frac{dy}{dx} = -\frac{1}{(1+x)\sqrt{1-x^2}}$

(10) $\frac{dy}{dx} = -\frac{|x|}{x^2\sqrt{x^2-1}}$

(11) $\frac{dy}{dx} = 2^x \ln 2$

(12) $\frac{dy}{dx} = 2xe^{x^2}$