Question

find dy/dx and d^2y/dx^2 if y=1+x/1+x^2

Answer 1

$y= \dfrac{1+x}{1+x^2}\,, \\ \,\\ \dfrac{dy}{dx} = \dfrac{(1+x)'(1+x^2) - (1+x)(1+x^2)'}{(1+x^2)^2} = \dfrac{(1+x^2) - (1+x)\cdot2x}{(1+x^2)^2} = \dfrac{1-2x-x^2}{(1+x^2)^2} \,, \\ \, \\ \dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \dfrac{1-2x-x^2}{(1+x^2)^2} = \dfrac{(1-2x-x^2)'(1+x^2)^2 - (1-2x-x^2)\big((1+x^2)^2\big)'}{(1+x^2)^4} = \dfrac{(-2-2x)(1+x^2)^2 - (1-2x-x^2)\cdot2(1+x^2)\cdot2x}{(1+x^2)^4} = \dfrac{2x^5+6x^4-4x^3+4x^2-6x-2}{(1+x^2)^4} = \dfrac{2(x-1)(x^2+1)(x^2+4x+1)}{(1+x^2)^4} = \dfrac{2(x-1)(x^2+4x+1)}{(1+x^2)^3}$