请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/04 18:23:09
请问设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0[sec（t）]^2d所确定,求d^2y/dx^2请问设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0[sec（t）]^
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
2x-tan(x-y)=∫(0,x-y) [sec（t）]^2dt
两边对x求导得：
2-sec²(x-y)(1-y')=sec²(x-y)(1-y')
sec²(x-y)(1-y')=1
y'=1-cos²(x-y)=sin²(x-y)
y''=2sin(x-y)cos(x-y)(1-y')
=sin(2x-2y)*cos²(x-y)
2x-tan(x-y)=∫(上限x-y下限0) [sec（t）]^2dt=tan(x-y)
=>x-tan(x-y)=0
=>dy/dx=1-sec²(x-y)
=>d^2y/dx^2=-2sec(x-y)tan(x-y)sec(x-y)=-2sec²(x-y)tan(x-y)
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec（t）]^2d所确定,求d^2y/dx^2
设函数y=y(x)由方程 y=tan(x+y) 所确定 求y''
设y=In(sec X+tan X ),求y'
设函数y=y(x)由方程lny=tan(xy)所确定,求dy
设y=In tan 2x 求微分dy
求证tan(x+y)*tan(x-y)=tan^2x-tan^2y/1-tan^2xtan^2y
解方程：dy/dx=y/x+tan(y/x)
设函数y=tan x,求y`=
tan(x+y)=
求证tan(x-y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-a)tan(z-x)
设y=(tan^2 x-csc^2 x )/(tan^2 x+cot^2 x -1) (a)证明y=1- 2/(tan^4 x-tan^2 x +1) (b)求y值的范围
已知方程y=tan(y-x)确定了y=y(x) 求dy
y(x)由方程y=tan(x+y)确定,求y''
设y=y(x)是由方程y=tan(x+y)所确定的隐函数,求微分dy如题
请问怎么证明sinX+sin(X+Y)+sin(X+2Y)/cosX+cos(X+Y)+cos(X+2Y)=tan(X+Y),
证明tan(x+y)+tan(x-y)=sin2x/cos^2x-sin^2y
y=tan 2x/tan x 的定义域是什么?y=tan 2x/tan x 的定义域是什么?
设tan[x+y]=5分之2,tan[y-4分之π,求]tan[x+4分之π的值]