设参数函数x=ln(1+t^2),y=t-arctant.求(d^2y)/(dx^2).另外,如果求dy/dx^2,d^2y/dx又怎样算?
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dy/dx=[1-1/(1+t²)] / [2t/(1+t²)]=t/2d²y/dx²=(1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1+t²)]=(1+t²)/(4t)希望可以帮到你,如果解决了问题,请点下面的选为满意回答按钮,...
======以下答案可供参考======
供参考答案1:
dx/dt=2t/(1+t^2)
dy/dt=1-1/(1+t^2)=t^2/(1+t^2)
dy/dx=(dy/dt)/(dx/dt)=t^2/(2t)=t/2
d^2y/dx^2=d(dy/dx)/dx=d(t/2)/dx=d(t/2)/dt /(dx/dt)=(1/2)/[2t/(1+t^2)]=(1+t^2)/(4t)