来自戴新的问题
设y=f(x)是由方程组x=3t^2+3y=te^t+1所确定后的隐函数,求f'(x)y=te^y+1
设y=f(x)是由方程组x=3t^2+3y=te^t+1所确定后的隐函数,求f'(x)
y=te^y+1
3回答
2020-05-21 10:17
设y=f(x)是由方程组x=3t^2+3y=te^t+1所确定后的隐函数,求f'(x)y=te^y+1
设y=f(x)是由方程组x=3t^2+3y=te^t+1所确定后的隐函数,求f'(x)
y=te^y+1
dx/dt=6t
dy/dt=e^t+te^t=(t+1)e^t
f'(x)=dy/dx=(dy/dt)/(dx/dt)=(t+1)e^t/(6t)
y=te^y+1题目错了,这样
y=te^y+1
对t求导:dy/dt=e^y+tdy/dt*e^y
得;dy/dt=e^y/(1-te^y)
因此f'(x)=e^y/[6t(1-te^y)]