What is the first and second derivative of #1/(x^2-x+2)#?

Question

3212 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-02T15:53:09

#f'(x)=(1-2x)/((x^2-x+2)^2)# and #f''(x)=(6x^2-6x-2)/((x^2-x+2)^3)#

Use the Quotient Rule #d/dx(g(x)/(h(x)))=(h(x)g'(x)-g(x)h'(x))/((h(x))^2)# to get

#f'(x)=(0*(x^2-x+2)-1*(2x-1))/((x^2-x+2)^2)=(1-2x)/((x^2-x+2)^2)#

Use the Quotient Rule again, along with the Chain Rule #d/dx(g(h(x))=g'(h(x))h'(x)# to get

#f''(x)=((x^2-x+2)^2*(-2) - (1-2x) * 2(x^2-x+2)*(2x-1))/((x^2-x+2)^4)#

Now simplify

#f''(x)=((x^2-x+2)(-2x^2+2x-4+2(2x-1)^2))/((x^2-x+2)^4)#

#=(-2x^2+2x-4+8x^2-8x+2)/((x^2-x+2)^3)=(6x^2-6x-2)/((x^2-x+2)^3)#