The derivative of #lnu#, by the chain rule, is #1/u*u'#. So we have to find the derivative of the inside function (which in our case is #(x)/(x-1)#) in order to find the derivative of the entire function. In math terms, we express this as:

#dy/dx=1/((x)/(x-1))*(x/(x-1))'#

#dy/dx=(x-1)/x*(x/(x-1))'#

Finding the derivative of #x/(x-1)# is going to be a little difficult because we need to use the quotient rule, which says:

#d/dx(u/v)=(u'v-uv')/v^2#, where (for our example) #u=x# and #v=x-1#.

Alright, let's get to work:

#d/dxx/(x-1)=((x)'(x-1)-(x)(x-1)')/(x-1)^2#

#d/dxx/(x-1)=(x-1-x)/(x-1)^2#

#d/dxx/(x-1)=(-1)/(x-1)^2#

Now we multiply this result by the derivative of #ln(x/(x-1))#, or #(x-1)/x#:

#(x-1)/x*(-1)/(x-1)^2#

Simplifying:

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

Or, equivalently,

#-1/(x^2-x)#