次の関数を微分せよ。 1) $3^x$ 2) $\frac{1}{\sqrt[5]{3x+1}}$ 3) $x^2 e^{-x}$ 4) $\sqrt{x^2+1}$ 5) $\log(x+\sqrt{x^2+1})$

1. 問題の内容

次の関数を微分せよ。

1) $3^x$

2) $\frac{1}{\sqrt[5]{3x+1}}$

3) $x^2 e^{-x}$

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

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

2. 解き方の手順

1) $3^x$ の微分

$y = 3^x$

$y' = 3^x \log 3$

2) $\frac{1}{\sqrt[5]{3x+1}}$ の微分

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

$y' = -\frac{1}{5} (3x+1)^{-\frac{6}{5}} \cdot 3 = -\frac{3}{5} (3x+1)^{-\frac{6}{5}} = -\frac{3}{5\sqrt[5]{(3x+1)^6}}$

3) $x^2 e^{-x}$ の微分

$y = x^2 e^{-x}$

$y' = (x^2)' e^{-x} + x^2 (e^{-x})' = 2x e^{-x} + x^2 (-e^{-x}) = 2x e^{-x} - x^2 e^{-x} = e^{-x}(2x-x^2)$

4) $\sqrt{x^2+1}$ の微分

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

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

5) $\log(x+\sqrt{x^2+1})$ の微分

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

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

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

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

3. 最終的な答え

1) $3^x \log 3$

2) $-\frac{3}{5\sqrt[5]{(3x+1)^6}}$

3) $e^{-x}(2x-x^2)$

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

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