1. ## The derivative..

Use the definition of the derivative to find f'(x)

f(x) = 1/(x-1)

I tried doing it, but have no idea how to do it.. I also tried the derivative "shortcut" and still got the wrong answer..

I tried using the shortcut and I ended up getting 0.. But how do they get -1/(x-1)^2??

2. Originally Posted by elpermic
Use the definition of the derivative to find f'(x)

f(x) = 1/(x-1)

I tried doing it, but have no idea how to do it.. I also tried the derivative "shortcut" and still got the wrong answer..

I tried using the shortcut and I ended up getting 0.. But how do they get -1/(x-1)^2??
you want

$\lim_{h \to 0} \frac {f(x + h) - f(x)}h = \lim_{h \to 0} \frac {\frac 1{x - 1 + h} - \frac 1{x - 1}}h$

or you may use the definition

$\lim_{x \to a} \frac {f(x) - f(a)}{x - a} = \lim_{x \to a} \frac {\frac 1{x - 1} - \frac 1{a - 1}}{x - a}$

in either case, combine the fractions in the numerator and simplify. you should be able to cancel the part in the denominator that gives zero when you try to take the limit.

we do the shortcut by the chain rule

we have $\frac d{dx} \bigg( \frac 1{x - 1} \bigg) = \frac d{dx}(x - 1)^{-1} = -(x - 1)^{-2}(1) = - \frac 1{(x - 1)^2}$