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

Dec 23, 2016

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

#### Explanation:

You could use the quotient rule, but I typically avoid doing this whenever possible as I find it tends to lead to higher chance of making an error and is generally more strenuous. To differentiate using the product rule, rewrite as

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

Product rule:

$f \left(x\right) = g \left(x\right) h \left(x\right)$

$f ' \left(x\right) = g \left(x\right) h ' \left(x\right) + g ' \left(x\right) h \left(x\right)$

In our case, $g \left(x\right) = x$ and $h \left(x\right) = {\left(x - 1\right)}^{-} 1$

Leaving $g \left(x\right)$ alone and multiply by the derivative of $h \left(x\right)$, for which we would use the chain rule.

$h ' \left(x\right) = - {\left(x - 1\right)}^{-} 2 \cdot 1$

Where $1$ is the derivative of the inside term, $x - 1$.

Then, we leave $h \left(x\right)$ alone and multiply by $g ' \left(x\right)$

$g ' \left(x\right) = 1$

Putting it all together, we have

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

Which is equivalent to

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

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

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