第3章 3.12

(1)

$y = \dfrac{1}{3x-2}$

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

$y'' = \dfrac{18}{(3x-2)^3}$

これから

$y^{(n)} = (-1)^n \dfrac{n! \cdot 3^n}{(3x-2)^{n+1}}$

(2)

$y = \dfrac{1+x}{1-x}$

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

$y'' = \dfrac{4}{(1-x)^3}$

これから

$y^{(n)} = \dfrac{2 \cdot n!}{(1-x)^{n+1}}$

(3)

$y = \dfrac{x^3}{1-x} = (-x^2-x-1) + \dfrac{1}{1-x}$

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

$y'' = -2+\dfrac{2}{(1-x)^3} \qquad y''' = \dfrac{6}{(1-x)^4}$

これから

$y^{(n)} = \dfrac{n!}{(1-x)^{n+1}} \qquad (n \ge 3)$

(4)

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

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

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

これから

$y^{(n)} = (-1)^n \frac{n!}{2}\left\{\frac{1}{(x-1)^{n+1}}+\frac{1}{(x+1)^{n+1}}\right\} \qquad (n \geq 2)$