Maths Help Evaluate f''(x)?

Sol
y=x^(3/2)/(x^2+1)
yx^2+y=x^(3/2)
2xy+x^2y’+y’=(3/2)x^(1/2)
(x^2+1)y’=(3/2)x^(1/2)-2xy
=(3/2)x^(1/2)－2x*x^(3/2)/(x^2+1)
=(3/2)x^(1/2)－2x^(5/2)/(x^2+1)
y’=(3/2)x^(1/2)/(x^2+1)－2x^(5/2)/(x^2+1)^2
(x^2+1)y’=(3/2)x^(1/2)-2xy
(x^2+1)y”+2xy’=(3/4)x^(－1/2)－2xy’－2y
(x^2+1)y”=(3/4)x^(－1/2)－4xy’－2y
=(3/4)x^(－1/2)－4x*[(3/2)x^(1/2)/(x^2+1)－2x^(5/2)/(x^2+1)^2]
－2x^(3/2)/(x^2+1)
=(3/4)x^(－1/2)－6x^(3/2)/(x^2+1)－8x^(7/2)/(x^2+1)^2－2x^(3/2)/(x^2+1)
=(3/4)x^(－1/2)－8x^(3/2)/(x^2+1)－8x^(7/2)/(x^2+1)^2
y”=(3/4)x^(－1/2)/(x^2+1)－8x^(3/2)/(x^2+1)^2－8x^(7/2)/(x^2+1)^3