Question

高等数学，第11题 求详解！

Answer 1

追答

可追问

追问

一阶导数那儿会了，第二部能在详细一些吗

追答

理解了吗？

谢谢

Answer 2

(11)
x=ln(1+t^2)
dx/dt = 2t/(1+t^2)

y=t-arcsint
dy/dt = 1- 1/√(1-t^2)

dy/dx
= dy/dt / dx/dt
=[1- 1/√(1-t^2)]/[2t/(1+t^2)]
=[ (1-t^2)-√(1-t^2)]/(2t)
= 1/(2t) - (1/2)t - √(1-t^2)]/(2t)

d/dt (dy/dx)
=-1/(2t^2) -(1/2) - [-t^2/√(1-t^2) -√(1-t^2) ] /(2t^2)
=-1/(2t^2) -(1/2) + 1/[2t^2.√(1-t^2)]

d^2y/dx^2
=d/dt (dy/dx) / dx/dt
={-1/(2t^2) -(1/2) + 1/[2t^2.√(1-t^2)]} /[2t/(1+t^2)]