设x=lncos t ,y=sin t - t*cos t 求d^2 y\(d x^2) 那个d^2 y dx^2 呢

Question

如题所述

Answer 1

d^2 y\(d x^2)=(d^2y\dxdt)*(dt\dx) dx=-tant dt dy=tsint dt dy\dx=-tcost (3)dt\dx=-1\tant 对(3)式微分d^2y\dxdt=-cost+tsint (d^2y\dxdt)*(dt\dx)=(-cost+tsint)\(-tant)