# Question #c5935

Mar 29, 2017

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

#### Explanation:

Write the function as:

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

Pose $y = \left(x - 1\right)$ and differentiate using the chain rule:

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

As:

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

we get:

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

Based on the power rule:

$\frac{d}{\mathrm{dy}} {y}^{n} = n {y}^{n - 1}$

we have:

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

And substituting $y = x - 1$ we can conclude:

$\frac{d}{\mathrm{dx}} \frac{1}{x - 1} = - \frac{1}{x - 1} ^ 2$