Question

1/1+e的x/6次方+e的x/3次方+e的x/2次方对x的微分

Answer 1

y=1/[1+e^(x/6)+e^(x/3)+e^(x/2)]

dy=(-1)* [(1/6)e^(x/6)+(1/3)e^(1/3)+(1/2)e^(x/2)]*(1 / [1+e^(x/6)+e^(x/3)+e^(x/2)]^2) *dx