数学凹凸性与增减性

y=1+36x/(x+3)^2

y'=[36(x^2+6x+9)-36x(2x+6)]/(x+3)^4
=[324-36x^2]/(x+3)^4
y''=[-72(x+3)^4-4(324-36x^2)(x+3)^3

y=x^2+1/x
y'=2x-1/x^2
y''=2-2/x^3

y=1+36x/(x+3)^2 ＝1＋36〔x/(x+3)^2〕
y'=36[(x^2+6x+9)-x(2x+6)]/(x+3)^4
=36[9-x^2]/(x+3)^4 ＝－36(x-3)/(x+3)^3
y''=-36{[(x+3)^3-3(x-3)(x+3)^2]/(x+3)^6}
=-36[x+3-3(x-3)]/(x+3)^4
=72(x-6)/(x+3)^4

y=x^2+1/x
y'=2x-1/x^2
y''=2＋2/x^3