How do you find the second derivative of ln(x^2-3x+3) ?

1 Answer
Nov 1, 2016

Firstly, use implicit differentiation...

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

e^y=x^2-3x+3

e^y*(dy)/(dx)=2x-3

(dy)/(dx)=(2x-3)/e^y

(dy)/(dx)=(2x-3)/(x^2-3x+3)

Now use the quotient rule...

(d^2y)/(dx^2)=((x^2-3x+3)*2-(2x-3)(2x-3))/((x^2-3x+3)^2)

And simplify the fraction...

(d^2y)/(dx^2)=(2x^2-6x+6-(4x^2-12x+9))/((x^2-3x+3)^2)

(d^2y)/(dx^2)=(2x^2-6x+6-4x^2+12x-9)/((x^2-3x+3)^2)

(d^2y)/(dx^2)=(-2x^2+6x-3)/((x^2-3x+3)^2)

(d^2y)/(dx^2)=(6x-2x^2-3)/((x^2-3x+3)^2)