問2.21の(1),(2),(3),(4)と問2.22の(1),(2),(3),(4)の関数の導関数をそれぞれ求めます。

1. 問題の内容

問2.21の(1),(2),(3),(4)と問2.22の(1),(2),(3),(4)の関数の導関数をそれぞれ求めます。

2. 解き方の手順

問2.21

(1) $y = x^{-\frac{4}{7}}$

$y' = -\frac{4}{7}x^{-\frac{4}{7} - 1} = -\frac{4}{7}x^{-\frac{11}{7}}$

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

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

(3) $y = 4\sqrt[3]{x^2} + 2 = 4x^{\frac{2}{3}} + 2$

$y' = 4 \cdot \frac{2}{3}x^{\frac{2}{3} - 1} = \frac{8}{3}x^{-\frac{1}{3}} = \frac{8}{3\sqrt[3]{x}}$

(4) $y = 5x^2\sqrt[5]{x^3} + 3 = 5x^2 x^{\frac{3}{5}} + 3 = 5x^{\frac{13}{5}} + 3$

$y' = 5 \cdot \frac{13}{5}x^{\frac{13}{5} - 1} = 13x^{\frac{8}{5}} = 13\sqrt[5]{x^8}$

問2.22

(1) $y = x^x$ ($x > 0$)

両辺の自然対数をとると $\ln y = \ln(x^x) = x \ln x$

両辺を $x$ で微分すると

$\frac{1}{y} \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

(2) $y = x^{\cos x}$ ($x > 0$)

両辺の自然対数をとると $\ln y = \cos x \ln x$

両辺を $x$ で微分すると

$\frac{1}{y} \frac{dy}{dx} = -\sin x \ln x + \cos x \cdot \frac{1}{x} = -\sin x \ln x + \frac{\cos x}{x}$

$\frac{dy}{dx} = y \left(-\sin x \ln x + \frac{\cos x}{x} \right) = x^{\cos x} \left(-\sin x \ln x + \frac{\cos x}{x} \right)$

(3) $y = (\log x)^x$ ($x > 1$)

両辺の自然対数をとると $\ln y = x \ln(\log x)$

両辺を $x$ で微分すると

$\frac{1}{y} \frac{dy}{dx} = \ln(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \ln(\log x) + \frac{1}{\log x}$

$\frac{dy}{dx} = y\left( \ln(\log x) + \frac{1}{\log x} \right) = (\log x)^x\left( \ln(\log x) + \frac{1}{\log x} \right)$

(4) $y = \frac{(x+1)^2}{(x+2)^3(x+3)^4}$

両辺の自然対数をとると

$\ln y = \ln((x+1)^2) - \ln((x+2)^3) - \ln((x+3)^4)$

$\ln y = 2\ln(x+1) - 3\ln(x+2) - 4\ln(x+3)$

両辺を $x$ で微分すると

$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+1} - \frac{3}{x+2} - \frac{4}{x+3}$

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

3. 最終的な答え

問2.21

(1) $y' = -\frac{4}{7}x^{-\frac{11}{7}}$

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

(3) $y' = \frac{8}{3\sqrt[3]{x}}$

(4) $y' = 13x^{\frac{8}{5}}$

問2.22

(1) $y' = x^x(\ln x + 1)$

(2) $y' = x^{\cos x} \left(-\sin x \ln x + \frac{\cos x}{x} \right)$

(3) $y' = (\log x)^x\left( \ln(\log x) + \frac{1}{\log x} \right)$

(4) $y' = \frac{(x+1)^2}{(x+2)^3(x+3)^4} \left( \frac{2}{x+1} - \frac{3}{x+2} - \frac{4}{x+3} \right)$