Answer
$F'(x)=4x+1+\dfrac{12}{x^3}$
Work Step by Step
$F(x)=\dfrac{2x^5+x^4-6x}{x^3}$
Use the quotient rule:
$\Bigg(\dfrac{f}{g}\Bigg)'=\dfrac{gf'-fg'}{g^2}$
so...
$F'(x)=\dfrac{(x^3)(10x^4+4x^3-6)-(2x^5+x^4-6x)(3x^2)}{x^6}$
Distribute:
$F'(x)=\dfrac{10x^7+4x^6-6x^3-6x^7-3x^6+18x^3}{x^6}$
Combine like terms:
$F'(x)=\dfrac{4x^7+x^6+12x^3}{x^6}$
Simplify:
$F'(x)=\dfrac{4x^7}{x^6}+\dfrac{x^6}{x^6}+\dfrac{12x^3}{x^6}$
$F'(x)=4x+1+\dfrac{12}{x^3}$
