y' = -2(x^-3)ln(x^2-3x+4) + (x^-2)(2x-3)(1/(x^2-3x+4))

y' = ( (-2/x)ln(x^2-3x+4) + (2x-3)/(x^2-3x+4) ) /x^2

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

- All categories
- Pre-Algebra Answers 12,306
- Algebra 1 Answers 25,235
- Algebra 2 Answers 10,430
- Geometry Answers 5,157
- Trigonometry Answers 2,637
- Calculus Answers 5,997
- Statistics Answers 3,016
- Word Problem Answers 10,085
- Other Math Topics 6,622

81,485 questions

85,664 answers

2,174 comments

69,138 users