Finding derivative of a fraction

5/((x^2) + x + 1)^2

i don't even know what rule to use...

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5/(x² + x + 1)² = 5(x² + x + 1) ⁻²            ← Now, take the derivative

d 5(x² + x + 1) ⁻²                                      d(x² + x + 1)
——————— = 5[-2(x² + x + 1) ⁻³] • ———————
          dx                                                        dx

                           = -10(x² + x + 1) ⁻³ • (2x+1)

                                 -10(2x+1)
                           = ——————            ← ANSWER
                               (x² + x + 1)³


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5/((x^2) + x + 1)^2

First multiply it out.

5/(x^4+2x^3+3x^2+2x+1)

Move x terms to top

5*(x^-4+1/2x^-3+1/3x^-2+1/2)

5x^-4+5/2x^-3+5/3x^-2+5/2

Now take derivative

-20x^-5+15/2x^-4+10/3x^-3

Simplify

-5x^-3 * (4x^-2+3/2x^-1+2/3)

if u and v are functions of x and y = u/v

dy/dx = [v(du/dx) - u(dv/dx)]/ v^2

your u = 5 so du/dx = 0

your v = ((x^2 + x + 1)^2 so dv/dx =2(x^2 + x + 1)(2x + 1)

v^2 = (x^2 + x + 1) ^4

Rewrite it as 5 * ((x^2) + x + 1)^-2