z'x=yx^(y-1),z'y=x^ylnx
x't=e^t,y't=1
dz/dt=z'x*x't+z'y*y't=yx^(y-1)e^t+x^ylnx最后答案是dz/dt=2te^(t^2)，是在你最后得出的式子上進(jìn)一步化簡(jiǎn)么？謝謝yx^(y-1)e^t+x^ylnx代入x=e^t，y=t=t(e^t)^(t-1)*e^t+(e^t)^t*t=te^(t^2)+e^(t^2)*t=2te^(t^2) 記得采納哦