sin+x+y+的二阶导数

权水妍1612matlab下求隐函数二阶导数如题sin(x+y)=x,写出程序 -
双转玉17562844070 ______[答案] clear all syms x y g=sym('sin(x+y(x))=x') dgdx2=diff(g,x,2)

权水妍1612y=sin(x+y),一阶隐导数y'=cos(x+y)/[1 - cos(x+y)]我懂,但如何求二阶隐导数~~~求过程求方法 -
双转玉17562844070 ______ 一阶导数你求对了 令 t = dy/dx = cos(x+y)/[1-cos(x+y)] dy = cos(x+y)·dx/[1-cos(x+y)] dx+dy = cos(x+y)·dx/[1-cos(x+y)] + dx = dx/[1-cos(x+y)] t+1 = 1+ cos(x+y)/[1-cos(x+y)] = 1/[1-cos(x+y)] 那么原函数的二阶导数即是 dt/dx t[1-cos(x+y)] = cos(x+y) [1-...

权水妍1612设函数y=y(x)满足方程e^y=sin(x+y).在点(∏/2)处,求二阶导数y＂和二阶微分dy -
双转玉17562844070 ______[答案] y=ln( sin(x+y) ) y'=[1/( sin(x+y) )]*cos(x+y)*(1+y') 解出y' y'=[1/( sin(x+y) )]*cos(x+y) / [1-[1/( sin(x+y) )]*cos(x+y)] 继续求导数 y''=[-y'[cos(x+y)(1+y')+sin(x+y)(1+y')]-cos(x+y)(1+y')]/[sin(x+y)-cos(x+y)] 代入点得到一阶导数,代入一阶导数得到二阶导数

权水妍1612设 x=ln(1+t^2),y=t - arctan(t),求d^3y/dx^3求方程y=sin(x+y)所确定的隐函数的二阶导数d^2y/dx^2求函数y=ln(1+x)/(1 - x)的n阶导数的一般表达式是y=ln[(1+x)/(1 - x)] -
双转玉17562844070 ______[答案] 设 x=ln(1+t^2),y=t-arctan(t),求d^3y/dx^3 dy/dt=1-1/(1+t^2) x=ln(1+t^2) 两边对x求导, 1=2tdt/dx/(1+t^2) dt/dx=(1+t^2)/2t dy/dx ... =(-16t^2-4)/16t^2*(1+t^2)/2t =-2(1+t^2)(1+4t^2)/16t^3 求方程y=sin(x+y)所确定的隐函数的二阶导数d^2y/dx^2 两边对x求导: ...

权水妍1612大学导数简单题求导.1.y=f(x^2)2.y=f(sin^2(x))+f(cos^2(x))3求二阶导,y=2x^2+lnx,二阶导数的意义是不是求两次导?我做出来y'=2x+1/x,y''=2+一串是错的, -
双转玉17562844070 ______[答案] y=f(x^2) y'=f'(x^2)*2x y=f(sin^2(x))+f(cos^2(x)) y'=f'(sin^2(x))*(sin^2x)'+f'(cos^2(x))*(cos^2x)' =f'(sin^2(x))*2sinccosx-f'(cos^2(x))*2sinccosx y=2x^2+lnx y'=4x+1/x y''=4-1/x^2

权水妍1612y=tan(x+y)的二级导数 -
双转玉17562844070 ______[答案] 二阶导数 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''=-2...

权水妍1612设y=x^2*f(sin x),求y的二阶导函数,其中f二阶可导设y=x²f(sin x),求y的二阶导函数,其中f二阶可导 -
双转玉17562844070 ______[答案] y'=2xf(sinx)+x^2cosxf'(sinx) y＂=2f(sinx)+2xcosxf'(sinx)+2xcosxf'(sinx)-x^2sinxf'(sinx)+x^2cos^2xf＂(sinx)

权水妍1612求某方程的二阶导数求由y=tan(x+y)所确定的隐函数y的二阶导数? -
双转玉17562844070 ______[答案] y=tan(x+y) =>arctany=x+y =>y=arctany-x (对方程两边 一阶求导) =>y'=y'/(1+y^2)-1 (对方程两边 二阶求导) =>y''=y''/(1+y^2)-2yy'/(1+y^2)^2

权水妍1612高数 求y=tan(x+y) 的二阶导数?谢谢 -
双转玉17562844070 ______[答案] y=tan(x+y)两边求导:y'=(sec(x+y))^2*(1+y'),所以y'=-(csc(x+y))^2=-1-1/y^2 所以,y''=2y'/y^3=-2(1+y^2)/y^5

权水妍1612y=cox(x+y)的二阶导数是 -
双转玉17562844070 ______[答案] 首先求一阶导数:y'=-sin(x+y)(x+y)'y'=-sin(x+y)(1+y') ====> y'=-sin(x+y)/(1+sin(x+y))上面式子两端继续对x求导:y''=-cos(x+y)(1+y') - sin(x+y)*y'' 乘积的求导法则整理有 y''=-cos(x+y)(1+y') / (1+sin(x+y))...