y=(1+x) / (3-x^2) 两个函数的商求导y' = [ (3-x^2) - (1+x)*(-2x) ] / (3-x^2)^2 = (x^2+2x+3) / (3-x^2)^2结果正确!
ax的倒数是ax^2的倒数是2x常数的倒数是0所以倒数应是:1/3-2x
y‘=2x+2+[(-3)/(3-x^2)^2]*[2(3-x^2)]*(-2x)=2x+2+12x/(3-x^2)]
y'=1/3-2x
f