ytanx+y的二阶导数

谷廖关3914已知Y=tan(X+Y),求Y的二阶导数. -
邵帘广15047428885 ______ y'=sec^2(x+y)*(1+y') 整理的y'=sec^2(x+y)/(1-sec^2(x+y))=-1/sin^2(x+y) 所以y''=(-1/sin^2(x+y))'=-2sin(x+y)cos(x+y)(1+y')/sin^4(x+y) =(2*sin(x+y)*cos(x+y)(1-1/sin^2(x+y)))/sin^4(x+y) =-2csc^2(x+y)*cot^3(x+y) 和上面的结果其实是一样的,我的更详细,哈哈

谷廖关3914y=tan(x+y) 求二阶导数 -
邵帘广15047428885 ______ y'= dy/dx =sec^2(x+y)·(1+y'); →[sec^2(x+y) -1]·y'=sec^2(x+y); →[tan^2(x+y) ]·y'=sec^2(x+y); →y'=1/sin^2(x+y); 则: y'' =dy' /dx =d[sin^(-2)(x+y)] /dx =(-2)·sin^(-3)(x+y) ·cos(x+y)·(1+y') =-2·sin^(-3)(x+y) ·cos(x+y)·[1+sin^(-2)(x+y)] =-2·cos(x+y)·[sin^(-3)(x+y) +sin^(-5)(x+y)]

谷廖关3914方程y=tan(x+y)所确定的函数的二阶导数 -
邵帘广15047428885 ______ y=tan(x+y) y'=sec²(x+y)*(x+y)' =sec²(x+y)*(1+y') =sec²(x+y)+y'sec²(x+y) y'-y'sec²(x+y)=sec²(x+y) y'=sec²(x+y)/[1-sec²(x+y)] =sec²(x+y)/{-[sec²(x+y)-1]} =sec²(x+y)/[-tan²(x+y)] =-1/cos²(x+y)*cos²(x+y)/sin²(x+y) =-csc²(x+y) y''=-2csc(x...

谷廖关3914求y=tan(x+y)的二阶导数 -
邵帘广15047428885 ______ y'= y'/[cos(x+y)]^2 y'[1-1/[cos(+y)]^2=0 y'*{-[tan(x+y)]^2=0 y'*(-y^2)=0 y^2y'=0 y'=0 y''=0

谷廖关3914大一高数:用对数法求y=tan(x+y)的二阶导数 -
邵帘广15047428885 ______ lny=ln tan(x+y) 两百求导 y'/y=(1+y')/tan(x+y)*sec^2(x+y) 解得 y'=y/(sec^2(x+y)/tan(x+y)-y)

谷廖关3914求隐函数y=tan(x+y)的二阶导数 -
邵帘广15047428885 ______ y=tan(x+y)得y'=(1+y')/(cos(x+y))^2 解得y'=-1/(sin(x+y))^2=-(sin(x+y))^(-2) y''=2(sin(x+y))^(-3)*cos(x+y)*(1+y') =2(sin(x+y))^(-3)*cos(x+y)*(1-1/(sin(x+y))^2) =-2(cos(x+y))^3(sin(x+y))^(-5)

谷廖关3914求详解y=tan(x+y)求隐函数的二阶导数d²y/dx² -
邵帘广15047428885 ______[答案] 你要求什么?dy/dx? 1*dy/dx= sec²(x+y) *(dy/dx+1) dy/dx(1-sec²(x+y))=sec²(x+y) dy/dx(-tan²(x+y))=sec²(x+y) dy/dx=-(1+tan²(x+y))/tan²(x+y) dy/dx=-cot²(x+y)-1 d²y/dx²=cot(x+y)csc(x+y)(1+dy/dx) =cot(x+y)csc(x+y)(-cot²(x+y)) =-cot³(x+y)csc(x...

谷廖关3914求下列方程所确定的隐函数y的二阶导数:y=tan(x+y) -
邵帘广15047428885 ______[答案] 本题运用隐函数求导法则和导数的四则运算,再进行代入即可求得答案:

谷廖关3914若由方程y=tan(x+y)所确定的隐函数为y=y(x),求y''(x)是二阶导数y''(x), -
邵帘广15047428885 ______[答案] y=tan(x+y)y'=sec²(x+y)*(x+y)'=sec²(x+y)*(1+y')=sec²(x+y)+y'sec²(x+y)y'-y'sec²(x+y)=sec²(x+y)y'=sec²(x+y)/[1-sec²(x+... (x+y)/{-[sec²(x+y)-1]}=sec²(x+y)/[-tan²(x+y)]=-1/cos²(x+y)*cos²(x+y)/sin²(x+y)=-csc²(x+y)y''=-2csc(x+y)*[-csc(x+y)cot(x+...

谷廖关3914求一个隐函数的二阶导数y=tan(x+y)求个详细步骤, -
邵帘广15047428885 ______[答案] 利用隐函数的微分法求令F(x,y(x))=0.两边对x求导,得:dF/dx+(dF/dy)*(dy/dx)=0.若dF/dy0,则dy/dx=-(dF/dx)/(dF/dy).于是题目可以这样设F=y-tan(x+y),dF/dx=-sec²(x+y),dF/dy=1-sec²(x+y)=-tan²(x...