How do you differentiate $f (x) = \frac{1}{x - 1}$ $f(x)=1/(x-1)$ ?

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Answer 1 · 2016-06-30T21:56:48

$= - \frac{1}{{(x - 1)}^{2}}$ $=-1/ (x-1) ^{2}$

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

so you could use the basic definition, namely that $d/dx ( x^n )= n x^{n-1} qquad square$

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

$f(u)=1/u$ where $u(x) = x-1$

and then we can say from the chain rule that

$(df)/dx = (df)/(du) * (du)/dx$

$= d/(du) (1/u) d/dx (x-1)$

$=color{blue}{ d/(du) (u^{-1})} d/dx (x-1)$

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

$color{blue}{-1 u ^{-2}} * 1$

$=-1/ u ^{2}$

$=-1/ (x-1) ^{2}$

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