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

Apr 26, 2016

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

Explanation:

We can use the Quotient Rule to find the derivative, which states:

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

Where in our case,

$h \left(x\right) = x$ and $g \left(x\right) = x - 1$

We then need to find the derivative of each of these:

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

Reassembling this we find our derivative:

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

Apr 26, 2016

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

Explanation:

In a previous answer, we used the quotient rule which is a simplification of the product rule. Some people, including me, don't like remembering more than one rule, so let's do that again using the product rule , which states:

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

in our case

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

Now we need to find the derivatives of these functions:

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

So reassembling we find our solution:

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