与えられた9個の関数を微分せよ。 (1) $\sqrt{\log x}$ (2) $\arctan(4x)$ (3) $(3x+1)^5$ (4) $7^x$ (5) $\frac{1}{\sqrt{2x^2+1}}$ (6) $x^2 \sin x$ (7) $(e^x+1)^3$ (8) $\tan(x^2+1)$ (9) $\arcsin(2x)$

1. 問題の内容

与えられた9個の関数を微分せよ。

(1) $\sqrt{\log x}$

(2) $\arctan(4x)$

(3) $(3x+1)^5$

(4) $7^x$

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

(6) $x^2 \sin x$

(7) $(e^x+1)^3$

(8) $\tan(x^2+1)$

(9) $\arcsin(2x)$

2. 解き方の手順

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

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

(2) $y = \arctan(4x)$

$\frac{dy}{dx} = \frac{1}{1+(4x)^2} \cdot 4 = \frac{4}{1+16x^2}$

(3) $y = (3x+1)^5$

$\frac{dy}{dx} = 5(3x+1)^4 \cdot 3 = 15(3x+1)^4$

(4) $y = 7^x$

$\frac{dy}{dx} = 7^x \log 7$

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

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

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

$\frac{dy}{dx} = 2x \sin x + x^2 \cos x$

(7) $y = (e^x+1)^3$

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

(8) $y = \tan(x^2+1)$

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

(9) $y = \arcsin(2x)$

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

3. 最終的な答え

(1) $\frac{1}{2x\sqrt{\log x}}$

(2) $\frac{4}{1+16x^2}$

(3) $15(3x+1)^4$

(4) $7^x \log 7$

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

(6) $2x \sin x + x^2 \cos x$

(7) $3e^x(e^x+1)^2$

(8) $2x \sec^2(x^2+1)$

(9) $\frac{2}{\sqrt{1-4x^2}}$