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

1 Answer
May 12, 2016

#(-1)/(x(x-1)#

Explanation:

Using #color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx(lnf(x))=1/(f(x)))color(white)(a/a)|)))#

combined with the #color(blue)" chain rule"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(d/dx[f(g(x))]=f'(g(x)).g'(x))color(white)(a/a)|)))color(green)" A"#
#"------------------------------------------------------------"#

here #f(g(x))=ln(x/(x-1))rArrf'(g(x))=1/(x/(x-1))=(x-1)/x#

and #g(x)=x/(x-1)rArrg'(x) = "see below"#
#"-----------------------------------------------------------"#
To differentiate g(x) we require to use the #color(blue)"quotient rule"#

If f(x)#=(g(x))/(h(x))" then " f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x)^2)color(green)"B"#
#"-----------------------------------------------------------------"#

here g(x) = x #rArrg'(x)=1#

and h(x) = x-1 #rArrh'(x)=1#
#"----------------------------------------------------------------"#
Substitute these values into #color(green)" B"#

#rArrf'(x)=((x-1).1-x.1)/(((x-1)^2))=(x-1-x)/(x-1)^2=(-1)/(x-1)^2#
#"------------------------------------------------------------"#
Now this is the value of g'(x) that we set out to obtain from above

Finally , 'plug' these all back into #color(green)"A"#

#rArrdy/dx=(x-1)/x xx(-1)/(x-1)^2#

#=cancel((x-1))/x xx(-1)/(cancel((x-1)) (x-1))=(-1)/(x(x-1))#