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
