Given: x^-2·ln(x^2-3x+4)
Differentiate the expression using the product rule: {f(x)g(x)}'=f'(x)g(x)+f(x)g'(x), getting
{(x^-2)ln(x^2-3x+4)}'=(x^-2)'{ln(x^2-3x+4)}+(x^-2){ln(x^2-3x+4)}'
Use the power rule:x^n=nx^(n-1), and a differentiation identity: (ln x)'=1/x, and the chain rule: {g(u(x))}'=g'(u(x))u'(x), u(x)=x^2-3x+4, getting
=(-2x^-3){ln(x^2-3x+4)}+(x^-2){(2x-3)/(x^2-3x+4)} Factor out by x^-3=1/x^3, getting
={-2·ln(x^2-3x+4)+x(2x-3)/(x^2-3x+4)}/x^3 Simplify, getting
={x·(2x-3)·(x2-3x+4)^-1 - 2·ln(x^2-3x+4)}/x^3
The answer is: d/dx{(x^-2)·ln(x^2-3x+4)} = {x·(2x-3)·(x^2-3x+4)^-1 - 2·ln(x^2-3x+4)}/x^3