We will be using the following properties of differentiation:

(power rule)

(1) #d/dx x^n = nx^(n-1)#

(2) #d/dxln(x) = 1/x#

(3) #d/dx f(kx) = kf'(x) " for " k in RR#

(4) #d/dx (f(x) + g(x)) = f'(x) + g'(x)#

(product rule)

(5) #d/dx f(x)g(x) = f(x)g'(x) + f'(x)g(x)#

Additionally, we will use the property of logarithms that

(6) #ln(x^n) = nln(x)#

With these in mind:

#d/dx (xln(x) - ln(x^2)) = xln(x) - 2ln(x)# (by 6)

#=> d/dx (xln(x) - ln(x^2)) = d/dxxln(x) - d/dx 2ln(x)# (by 4)

Solving for the first term:

#d/dxxln(x) = x(d/dxln(x)) + (d/dxx)ln(x)# (by 5)

#=> d/dx xln(x) = x(1/x) + 1*ln(x)# (by 1 and 2)

#=> d/dx xln(x) = 1 + ln(x)#

Solving for the second term:

#d/dx 2ln(x) = 2(d/dx ln(x)# (by 3)

#=> d/dx 2ln(x) = 2(1/x) = 2/x# (by 2)

Thus, substituting,

#d/dx (xln(x) - ln(x^2)) = 1 + ln(x) - 2/x#