y= x^-2 ln (x^2-3x+4)
asked Oct 6, 2014 in Calculus Answers by anonymous

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2 Answers

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

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
answered Oct 6, 2014 by johnperkins Top Rated User (103,080 points)

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

answered Oct 6, 2014 by tokiokid70
edited Oct 6, 2014
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