# What is the second derivative of the function f(x) = (x) / (x - 1)?

Oct 6, 2016

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \frac{x}{x - 1} = \frac{2}{x - 1} ^ 3$

#### Explanation:

For this problem, we will use the quotient rule:

$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

We can also make it a little easier by dividing to get

$\frac{x}{x - 1} = 1 + \frac{1}{x - 1}$

First derivative:

$\frac{d}{\mathrm{dx}} \left(1 + \frac{1}{x - 1}\right)$

$= \left(\frac{d}{\mathrm{dx}} 1\right) + \left(\frac{d}{\mathrm{dx}} \frac{\left(x - 1\right) \left(\frac{d}{\mathrm{dx}} 1\right) - 1 \left(\frac{d}{\mathrm{dx}} \left(x - 1\right)\right)}{x - 1} ^ 2\right)$

$= 0 + \frac{\left(x - 1\right) \left(0\right) - \left(1\right) \left(1\right)}{x - 1} ^ 2$

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

Second derivative:

The second derivative is the derivative of the first derivative.

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \left(1 + \frac{1}{x - 1}\right) = \frac{d}{\mathrm{dx}} \left(- \frac{1}{x - 1} ^ 2\right)$

$= - \frac{{\left(x - 1\right)}^{2} \left(\frac{d}{\mathrm{dx}} 1\right) - 1 \left(\frac{d}{\mathrm{dx}} {\left(x - 1\right)}^{2}\right)}{{\left(x - 1\right)}^{2}} ^ 2$

$= - \frac{{\left(x - 1\right)}^{2} \left(0\right) - 1 \left(2 \left(x - 1\right)\right)}{x - 1} ^ 4$

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

We could also have used the power rule $\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$ for $n \ne 1$:

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

$\implies \frac{d}{\mathrm{dx}} \left(1 + \frac{1}{x - 1}\right) = \frac{d}{\mathrm{dx}} \left(1 + {\left(x - 1\right)}^{- 1}\right)$

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

$\implies {d}^{2} / \left({\mathrm{dx}}^{2}\right) \left(1 + \frac{1}{x - 1}\right) = \frac{d}{\mathrm{dx}} \left(- {\left(x - 2\right)}^{- 2}\right)$

$= 2 {\left(x - 2\right)}^{- 3}$

which is the same as the result we obtained above.