第6章 6.10

(1)

両辺対数をとり

$\log y = \sqrt{x} \log a$

$\frac{1}{y}y' = \frac{1}{2\sqrt{x}}\log a \qquad y' = \frac{y}{2\sqrt{x}}\log a = \frac{a^{\sqrt{x}}}{2\sqrt{x}}\log a$

(2)

両辺対数をとり

$\log y = \sin x \cdot \log(\cos x)$

$\frac{1}{y}y' = \cos x \cdot \log(\cos x) + \sin x \cdot \frac{-\sin x}{\cos x} = \cos x \cdot \log(\cos x) - \sin x \cdot \tan x$

$y' = y\left(\cos x \cdot \log(\cos x) - \sin x \cdot \tan x\right) = \left(\cos x \cdot \log(\cos x) - \sin x \cdot \tan x\right)(\cos x)^{\sin x}$

(3)

両辺対数をとり

$\log y = \log x \cdot \log x$

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

(4)

両辺対数をとり

$\log y = \sqrt{x}\log x$

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