第1个回答 2019-09-19
y=√[xsinx .√(1-e^x)]
lny =(1/2)lnx +(1/2)lnsinx + (1/4)ln(1-e^x)
dy/y =(1/2)(1/x)dx +(1/2)cotx.dx + (1/4)[1/(1-e^x)] [ -e^x .dx]
={(1/2)(1/x) +(1/2)cotx - (1/4)[e^x/(1-e^x)] } dx
dy =√[xsinx .√(1-e^x)] .{(1/2)(1/x) +(1/2)cotx - (1/4)[e^x/(1-e^x)] } dx追问
lny =(1/2)lnx +(1/2)lnsinx + (1/4)ln(1-e^x)
dy/y =(1/2)(1/x)dx +(1/2)cotx.dx + (1/4)[1/(1-e^x)] [ -e^x .dx]
={(1/2)(1/x) +(1/2)cotx - (1/4)[e^x/(1-e^x)] } dx
dy =√[xsinx .√(1-e^x)] .{(1/2)(1/x) +(1/2)cotx - (1/4)[e^x/(1-e^x)] } dx追问
写掉了一个1/x是吧
本回答被提问者采纳
相似回答