Question

过程求解，谢谢

Answer 1

=lime^(2xln(1+1/x))=e^2

追答

=lim5x/2x=5/2

=-tan(e^x)e^x

y′sinx+ycosx+sin(x+y)(1+y′)=0
y′=-(ycosx+sin(x+y))/(sinx+sin(x+y))

追问

能不能给我详细过程呢？

追答

…这已经是详细过程了，除非你一点基础都没有。

Answer 2

如图

Answer 3

(1)lim (1+1/x)^(2x)=lim[ (1+1/x)^x]²=e²
x→∞
(2)lim tan5x/2x=5x/2x=5/2
x→0
(3)y'={ ln[cos(e^x)] }'=[cos(e^x)]'/cos(e^x)=-(e^x)'sin(e^x)/cos(e^x)
y'=-e^xtan(e^x)
(4) ysinx-cos(x+y)=0
sinxdy+ycosxdx+sin(x+y)(dx+dy)=0
[sinx+sin(x+y)]dy=-[ycosx+sin(x+y)]dx
dy/dx=-[ycosx+sin(x+y)]/(sinx+sin(x+y))
dy/dx|(0,π/2)=1-π/2

Answer 4

给点财富我用用可以吗

追问

你要财富干嘛？你可以做任务啊，挣啊