与えられた15個の関数を微分する問題です。

1. 問題の内容

与えられた15個の関数を微分する問題です。

2. 解き方の手順

微分公式を用いて各関数を微分します。

(1) $y = x^4$

$y' = 4x^{4-1} = 4x^3$

(2) $y = x^6$

$y' = 6x^{6-1} = 6x^5$

(3) $y = x^{100}$

$y' = 100x^{100-1} = 100x^{99}$

(4) $y = \sqrt{x} = x^{1/2}$

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

(5) $y = \sqrt[4]{x^3} = x^{3/4}$

$y' = \frac{3}{4}x^{\frac{3}{4}-1} = \frac{3}{4}x^{-\frac{1}{4}} = \frac{3}{4\sqrt[4]{x}}$

(6) $y = \frac{1}{x} = x^{-1}$

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

(7) $y = \frac{1}{x^2} = x^{-2}$

$y' = -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3}$

(8) $y = \frac{1}{x^4} = x^{-4}$

$y' = -4x^{-4-1} = -4x^{-5} = -\frac{4}{x^5}$

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

$y' = -\frac{1}{3}x^{-\frac{1}{3}-1} = -\frac{1}{3}x^{-\frac{4}{3}} = -\frac{1}{3x\sqrt[3]{x}}$

(10) $y = \frac{1}{x\sqrt{x}} = \frac{1}{x^{3/2}} = x^{-3/2}$

$y' = -\frac{3}{2}x^{-\frac{3}{2}-1} = -\frac{3}{2}x^{-\frac{5}{2}} = -\frac{3}{2x^2\sqrt{x}}$

(11) $y = \sin x$

$y' = \cos x$

(12) $y = \cos x$

$y' = -\sin x$

(13) $y = 2^x$

$y' = 2^x \ln 2$

(14) $y = e^x$

$y' = e^x$

(15) $y = \log x$ (底がe、つまり自然対数)

$y' = \frac{1}{x}$

3. 最終的な答え

(1) $y' = 4x^3$

(2) $y' = 6x^5$

(3) $y' = 100x^{99}$

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

(5) $y' = \frac{3}{4\sqrt[4]{x}}$

(6) $y' = -\frac{1}{x^2}$

(7) $y' = -\frac{2}{x^3}$

(8) $y' = -\frac{4}{x^5}$

(9) $y' = -\frac{1}{3x\sqrt[3]{x}}$

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

(11) $y' = \cos x$

(12) $y' = -\sin x$

(13) $y' = 2^x \ln 2$

(14) $y' = e^x$

(15) $y' = \frac{1}{x}$