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#
