急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2
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急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2
![急设x=2t^(2)-1,y=根号(1+t^2).求dy/dx和d^2y/dx^2](/uploads/image/z/389586-66-6.jpg?t=%E6%80%A5%E8%AE%BEx%3D2t%5E%282%29-1%2Cy%3D%E6%A0%B9%E5%8F%B7%281%2Bt%5E2%29.%E6%B1%82dy%2Fdx%E5%92%8Cd%5E2y%2Fdx%5E2)
dy/dx=(dy/dt)/(dx/dt)=[t/√(1+t^2)]/(4t)=1/[4√(1+t^2)]
d^2y/dx^2=[d(dy/dx)/dt]/dx/dt
=1/4*(-1/2)*(1+t^2)^(-3/2)/(4t)
=-1/32*(1+t^2)^(-3/2)/t
d^2y/dx^2=[d(dy/dx)/dt]/dx/dt
=1/4*(-1/2)*(1+t^2)^(-3/2)/(4t)
=-1/32*(1+t^2)^(-3/2)/t
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