Question

x=f'(t),y=tf'(t)-f(t),f''(t)存在且不为0,求d2y/dx2.这里没有复合求导的问题吗?怎么就

Answer 1

当然是要复合求导的,

x=f'(t),y=tf'(t)-f(t)

所以

dx/dt=f "(t)

dy/dt=t *f "(t) +f '(t)- f '(t)=t *f "(t),

所以

dy/dx=(dy/dt)/(dx/dt)=t,

而

d²y/dx²

=d(dy/dx) /dx

= d(dy/dx) /dt * dt/dx

= dt/dt * dt/dx

=1/(dx/dt)

=1/f "(t)

所以y对x的二阶导数d²y/dx²=1/f "(t)