How do you differentiate #y=(5-2x+3x^2)/(x^2+3)#?

1 Answer
Nov 6, 2016

#\frac{dy}{dx}=\frac{2(x^2+4x-3)}{(x^2+3)^2}#

Explanation:

The quotient rule says:
if #y=\frac{a}{b}# then #\frac{dy}{dx}=\frac{b\cdot\frac{d}{dx}[a]-a\cdot\frac{d}{dx}[b]}{b^2}#
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}#
calculate the derivatives
#\frac{dy}{dx}=\frac{(x^2+3)(6x-2)-(5-2x+3x^2)(2x)}{(x^2+3)^2}#
expand
#\frac{dy}{dx}=\frac{(6x^3+18x-2x^2-6)-(10x-4x^2+6x^3)}{(x^2+3)^2}#

more expanding
#\frac{dy}{dx}=\frac{6x^3+18x-2x^2-6-10x+4x^2-6x^3}{(x^2+3)^2}#

group like terms
#\frac{dy}{dx}=\frac{6x^3-6x^3+18x-10x-2x^2+4x^2-6}{(x^2+3)^2}#

simplify
#\frac{dy}{dx}=\frac{8x+2x^2-6}{(x^2+3)^2}#

#\frac{dy}{dx}=\frac{2(x^2+4x-3)}{(x^2+3)^2}#