$a)y=\dfrac{4x^3+2x^2+1}{x^2+2}\\ y'=\dfrac{(4x^3+2x^2+1)'(x^2+2)-(4x^3+2x^2+1)(x^2+2)'}{(x^2+2)^2}\\ = \dfrac{(12x^2+4x)(x^2+2)-(4x^3+2x^2+1).2x}{(x^2+2)^2}\\ = \dfrac{4x^4+24x^2+6x}{(x^2+2)^2}\\ b)y=\sqrt{3x^4+4x^2+2}\\ y'=\dfrac{1}{2\sqrt{3x^4+4x^2+2}}.(3x^4+4x^2+2)'\\ =\dfrac{1}{2\sqrt{3x^4+4x^2+2}}.(12x^3+8x)\\ =\dfrac{6x^3+4x}{\sqrt{3x^4+4x^2+2}}\\ c)y=3\sin(x+2)+\cos\left(x-\dfrac{\pi}{3}\right)\\ =3\cos(x+2)-\sin\left(x-\dfrac{\pi}{3}\right)\\ d)y=\dfrac{x+2}{\sqrt{x^2+4}}\\ y'=\dfrac{(x+2)'\sqrt{x^2+4}-(x+2)\sqrt{x^2+4}'}{x^2+4}\\ =\dfrac{\sqrt{x^2+4}-(x+2)\dfrac{(x^2+4)'}{2\sqrt{x^2+4}}}{x^2+4}\\ =\dfrac{\sqrt{x^2+4}-(x+2)\dfrac{2x}{2\sqrt{x^2+4}}}{x^2+4}\\ =\dfrac{\sqrt{x^2+4}-\dfrac{x(x+2)}{\sqrt{x^2+4}}}{x^2+4}\\ =\dfrac{\dfrac{x^2+4}{\sqrt{x^2+4}}-\dfrac{x(x+2)}{\sqrt{x^2+4}}}{x^2+4}\\ =\dfrac{\dfrac{x^2+4-x(x+2)}{\sqrt{x^2+4}}}{x^2+4}\\ =\dfrac{4-2x}{\sqrt{(x^2+4)^3}}$
