Question

y=tan（x+y) 求二阶导数

Answer 1

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)]