第6章 6.15

(1)

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

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

(2)

$y = \log(\cos x)$

$y' = -\frac{\sin x}{\cos x} = -\tan x$

(3)

$y = \frac{a^x}{x^2+a^2}$

公式より $(a^x)' = a^x \log a$

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

(4)

$y = e^{-x}\sin x$

$y' = -e^{-x}\sin x + e^{-x}\cos x = e^{-x}(\cos x - \sin x)$

(5)

$y = x\sqrt{1+x^2} + \log\left|x+\sqrt{1+x^2}\right|$

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

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

$= \sqrt{1+x^2} + \sqrt{1+x^2} = 2\sqrt{1+x^2}$

(6)

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

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

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

$= \frac{1}{\sqrt{x+2}\sqrt{x+1}} = \frac{1}{\sqrt{x^2+3x+2}}$