高数的解答题...,x=ln(1+t^2) y=t-arctan t 求dy/dx

高数的解答题...,x=ln(1+t^2) y=t-arctan t 求dy/dx
高数的解答题...,x=ln(1+t^2) y=t-arctan t 求dy/dx

高数的解答题...,x=ln(1+t^2) y=t-arctan t 求dy/dx
x=ln(1+t^2)
dx/dt=2t/(1+t²)
y=t-arctan t
dy/dt=1-1/(1+t²)=t²/(1+t²)
所以
dy/dx=t²/(1+t²)/[2t/(1+t²)]=t/2

dy/dt=1-1/(1+t²)=t²/(1+t²)
dx/dt=2t/(1+t²)
dy/dx
=(dy/dt)/(dx/dt)
=t/2

dy/dx=（dy/dt)/(dx/dt)
dy/dt=1-1/(1+t²)=t²/(1+t²)
dx/dt=2t/(1+t²)
dy/dx=t/2