# How do you find the derivative of y= ln (x/(x-1))?

Feb 12, 2016

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

#### Explanation:

The derivative of $\ln u$, by the chain rule, is $\frac{1}{u} \cdot u '$. So we have to find the derivative of the inside function (which in our case is $\frac{x}{x - 1}$) in order to find the derivative of the entire function. In math terms, we express this as:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{x}{x - 1}} \cdot \left(\frac{x}{x - 1}\right) '$

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

Finding the derivative of $\frac{x}{x - 1}$ is going to be a little difficult because we need to use the quotient rule, which says:
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{u ' v - u v '}{v} ^ 2$, where (for our example) $u = x$ and $v = x - 1$.
Alright, let's get to work:
$\frac{d}{\mathrm{dx}} \frac{x}{x - 1} = \frac{\left(x\right) ' \left(x - 1\right) - \left(x\right) \left(x - 1\right) '}{x - 1} ^ 2$
$\frac{d}{\mathrm{dx}} \frac{x}{x - 1} = \frac{x - 1 - x}{x - 1} ^ 2$
$\frac{d}{\mathrm{dx}} \frac{x}{x - 1} = \frac{- 1}{x - 1} ^ 2$

Now we multiply this result by the derivative of $\ln \left(\frac{x}{x - 1}\right)$, or $\frac{x - 1}{x}$:
$\frac{x - 1}{x} \cdot \frac{- 1}{x - 1} ^ 2$
Simplifying:
$- \frac{1}{x \left(x - 1\right)}$
Or, equivalently,
$- \frac{1}{{x}^{2} - x}$