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

1 Answer
Apr 25, 2017

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

Explanation:

differentiate using the #color(blue)"quotient rule"#

#"Given " f(x)=(g(x))/(h(x))" then"#

#f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"#

#"here " g(x)=xrArrg'(x)=1#

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

#color(orange)"Reminder " d/dx(ln(f(x)))=(f'(x))/((f(x))#

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

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

#color(white)(rArrf'(x))=(1-ln(x-1)+x/(x-1))/(1-ln(x-1))^2#