The quotient rule says:
if y=\frac{a}{b}y=ab then \frac{dy}{dx}=\frac{b\cdot\frac{d}{dx}[a]-a\cdot\frac{d}{dx}[b]}{b^2}dydx=b⋅ddx[a]−a⋅ddx[b]b2
so
\frac{dy}{dx}=\frac{(x^2+3)\cdot\frac{d}{dx}(5-2x+3x^2)-(5-2x+3x^2)\cdot\frac{d}{dx}(x^2+3)}{(x^2+3)^2}dydx=(x2+3)⋅ddx(5−2x+3x2)−(5−2x+3x2)⋅ddx(x2+3)(x2+3)2
calculate the derivatives
\frac{dy}{dx}=\frac{(x^2+3)(6x-2)-(5-2x+3x^2)(2x)}{(x^2+3)^2}dydx=(x2+3)(6x−2)−(5−2x+3x2)(2x)(x2+3)2
expand
\frac{dy}{dx}=\frac{(6x^3+18x-2x^2-6)-(10x-4x^2+6x^3)}{(x^2+3)^2}dydx=(6x3+18x−2x2−6)−(10x−4x2+6x3)(x2+3)2
more expanding
\frac{dy}{dx}=\frac{6x^3+18x-2x^2-6-10x+4x^2-6x^3}{(x^2+3)^2}dydx=6x3+18x−2x2−6−10x+4x2−6x3(x2+3)2
group like terms
\frac{dy}{dx}=\frac{6x^3-6x^3+18x-10x-2x^2+4x^2-6}{(x^2+3)^2}dydx=6x3−6x3+18x−10x−2x2+4x2−6(x2+3)2
simplify
\frac{dy}{dx}=\frac{8x+2x^2-6}{(x^2+3)^2}dydx=8x+2x2−6(x2+3)2
\frac{dy}{dx}=\frac{2(x^2+4x-3)}{(x^2+3)^2}dydx=2(x2+4x−3)(x2+3)2
