# How do you differentiate f(x)=1/(x-1)?

Jun 30, 2016

$= - \frac{1}{x - 1} ^ \left\{2\right\}$

#### Explanation:

in any number of ways. it is $f \left(x\right) = {\left(x - 1\right)}^{- 1}$

so you could use the basic definition, namely that $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1} q \quad \square$

but here it is (x-1) and not x so we might wish to look at the chain rule and an intermediate substitution

we can say that

$f \left(u\right) = \frac{1}{u}$ where $u \left(x\right) = x - 1$

and then we can say from the chain rule that

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

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

$= \textcolor{b l u e}{\frac{d}{\mathrm{du}} \left({u}^{- 1}\right)} \frac{d}{\mathrm{dx}} \left(x - 1\right)$

the blue bit calls on the idea in $\square$ and we get

$\textcolor{b l u e}{- 1 {u}^{- 2}} \cdot 1$

$= - \frac{1}{u} ^ \left\{2\right\}$

$= - \frac{1}{x - 1} ^ \left\{2\right\}$