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

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Answer 1 · 2018-04-12T19:20:41

#f'(x)=(1-ln(x-1)+1/(x-1))/(1-ln(x-1))^2#

#(d(g(x))/(h(x)))/(dx)=(g'(x)h(x)-g(x)h'(x))/(h(x))^2#

Here

#g(x)=x#

#g'(x)=1#

#h(x)=1-ln(x-1)#, and

#h'(x)=-1/(x-1)#

So

#f'(x)=(1-ln(x-1)-(1)(-1/(x-1)))/(1-ln(x-1))^2#

#f'(x)=(1-ln(x-1)+1/(x-1))/(1-ln(x-1))^2#