已知x=arctant+sint,t+y-e^(yt)=0,求当t=0时该曲线的切线方程.
设F(y,t)=t+y-e^(yt)=0
那么dy/dt=-(∂F/∂t)/(∂F/∂y)=-[1-ye^(yt)]/[1-te^(yt)],故：
dy/dx=(dy/dt)/(dx/dt)={-[1-ye^(yt)]/[1-te^(yt)]}/[1/(1+t²)+cost]
={-[1-ye^(yt)](1+t²)}/{[1-te^(yt)][1+(1+t²)cost]}
当t=0时x=0,y=1,代入上式,即得t=0时dy/dx=0
故当t=0是曲线的切线方程为y=1.