Answer
$f''(x)=\dfrac{2}{(x-1)^3}$
Work Step by Step
First Derivative:
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=x; u'(x)=1$
$v(x)=x-1; v'(x)=1$
$f'(x)=\frac{(1)(x-1)-(1)(x)}{(x-1)^2}=-\frac{1}{(x-1)^2}.$
Second Derivative:
Using the quotient rule: $fâ'(x)=\frac{d}{dx}f'(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=-1; u'(x)=0$
$v(x)=(x-1)^2; v'(x)=2(x-1)$
$f''(x)=\frac{(0)(x-1)^2-(-1)(2x-2)}{((x-2)^2)^2}=\dfrac{2}{(x-1)^3}.$
