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)]

Question

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)]

Answer 1

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
=dy/dt*dt/dx
=[1-1/(1+t^2)](1+t^2)/2t
=t/2
d^2y/dx^2=1/2dt/dx=(1+t^2)/4t
d^3y/dx^3
=[(1+t^2)'4t-(1+4t^2)(4t)']/(16t^2)
=[8t^2dt/dx-4(1+4t^2)dt/dx]/16t^2
=(-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:
dy/dx=cos(x+y)(1+dy/dx)
dy/dx=cos(x+y)/[1-cos(x+y)]
再对x求导:
d2y/dx^2=sin(x+y)(1+dy/dx)(1+dy/dx)+cos(x+y)d^2y/dx^2
d2y/dx^2=sin(x+y)(1+dy/dx)^2/cos(x+y)
=tan(x+y)(1+cos(x+y)/[1-cos(x+y)]
求函数y=ln(1+x)/(1-x)的n阶导数的一般表达式
[(1+x)/(1-x)]'=2/(1-x)^2
y'=1/(x+1)-1/(x-1)
y''=-1/(x+1)^2+1/(x-1)^2
y(3)=2/(x+1)^3-2/(x-1)^3
...
y(n)=(-1)^(n-1)*(n-1)!/(x+1)^n+(-1)^n*(n-1)!/(x-1)^n