What is the second derivative of x/(x-1) and the first derivative of 2/x?

Apr 7, 2015

Question 1
If $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$ then by the Quotient Rule

$f ' \left(x\right) = \frac{g ' \left(x\right) \cdot h \left(x\right) - g \left(x\right) \cdot h ' \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$

So if $f \left(x\right) = \frac{x}{x - 1}$
then the first derivative
$f ' \left(x\right) = \frac{\left(1\right) \left(x - 1\right) - \left(x\right) \left(1\right)}{x} ^ 2$

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

and the second derivative is
$f ' ' \left(x\right) = 2 {x}^{-} 3$

Question 2
If $f \left(x\right) = \frac{2}{x}$ this can be re-written as

$f \left(x\right) = 2 {x}^{-} 1$

and using standard procedures for taking the derivative
$f ' \left(x\right) = - 2 {x}^{-} 2$
or, if you prefer
$f ' \left(x\right) = - \frac{2}{x} ^ 2$