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

Question

22346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-30T21:56:48

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

in any number of ways. it is #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}#