derivative

d/dx(lnx)=1/x 

d/dx(x)=1 

the first part is a product rule so f'(x)g(x)+f(x)g'(x)

f(x)=x

g(x)=ln(x)

ln(x)+x(1/x)-1

simplify it and;

ln(x)-1

Possible derivation:
d/dx(x log(x)-x)
Differentiate the sum term by term and factor out constants:
  =  -(d/dx(x))+d/dx(x log(x))
The derivative of x is 1:
  =  d/dx(x log(x))-1
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = log(x):
  =  -1+log(x) d/dx(x)+x d/dx(log(x))
Simplify the expression:
  =  -1+log(x) (d/dx(x))+x (d/dx(log(x)))
The derivative of x is 1:
  =  -1+x (d/dx(log(x)))+1 log(x)
The derivative of log(x) is 1/x:
  =  -1+log(x)+1/x x
Simplify the expression:
Answer: | =  log(x)