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

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Answer 1 · 2016-11-01T18:41:42

#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)#