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

1 Answer
Jul 2, 2015

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

Explanation:

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)