역함수 미분

$\displaystyle f' ( x ) = \frac{e ^{x}}{e ^{x} - 1}$ $\cdots \cdots$$\cdots \cdots$$\cdots \cdots$$\cdots \cdots$$\cdots \cdots$(ⅰ) $y = g ( x )$ 에서, $x = f ( y ) = \ln ( e ^{y} - 1 )$ $\therefore e ^{y} - 1 = e ^{x}$ $\displaystyle g' ( x ) = \frac{1}{f' ( y )} = \frac{e ^{y} - 1}{e ^{y}} = \frac{e ^{x}}{e ^{x} + 1}$ $\cdots \cdots$$\cdots \cdots$$\cdots \cdots$(ⅱ)

(ⅰ), (ⅱ) 에 의해 $\displaystyle \frac{1}{f ' ( a )} + \frac{1}{g ' ( a )} = \frac{e ^{a} - 1}{e ^{a}} + \frac{e ^{a} + 1}{e ^{a}} = 2$