次の関数を微分せよ。 (1) $f(x) = \frac{\sqrt{x}+2}{\sqrt{x}+2}$ (2) $f(x) = \frac{1}{\sqrt{x^2+1}}$ (3) $f(x) = x\sqrt{a^2-x^2} + a^2\arcsin\frac{x}{a}$ ($a>0$) (4) $f(x) = \log|x+\sqrt{x^2+a}|$

はい、承知いたしました。問題を解いていきましょう。

1. 問題の内容

次の関数を微分せよ。

(1) $f(x) = \frac{\sqrt{x}+2}{\sqrt{x}+2}$

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

(3) $f(x) = x\sqrt{a^2-x^2} + a^2\arcsin\frac{x}{a}$ ($a>0$)

(4) $f(x) = \log|x+\sqrt{x^2+a}|$

2. 解き方の手順

(1)

$f(x) = \frac{\sqrt{x}+2}{\sqrt{x}+2} = 1$

したがって、$f'(x) = 0$

(2)

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

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

(3)

$f(x) = x\sqrt{a^2-x^2} + a^2\arcsin\frac{x}{a}$

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

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

$f'(x) = \frac{a^2-x^2-x^2}{\sqrt{a^2-x^2}} + \frac{a}{\sqrt{\frac{a^2-x^2}{a^2}}}$

$f'(x) = \frac{a^2-2x^2}{\sqrt{a^2-x^2}} + \frac{a^2}{\sqrt{a^2-x^2}}$

$f'(x) = \frac{2a^2-2x^2}{\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{\sqrt{a^2-x^2}} = 2\sqrt{a^2-x^2}$

(4)

$f(x) = \log|x+\sqrt{x^2+a}|$

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

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

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

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

3. 最終的な答え

(1) $0$

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

(3) $2\sqrt{a^2-x^2}$

(4) $\frac{1}{\sqrt{x^2+a}}$