x=ln(1+t^2),y=arctant 求dy/dx,和d2y/dx2

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x=ln(1+t^2),y=arctant 求dy/dx,和d2y/dx2

x=ln(1+t^2),y=arctant 求dy/dx,和d2y/dx2
x=ln(1+t^2),y=arctant 求dy/dx,和d2y/dx2

x=ln(1+t^2),y=arctant 求dy/dx,和d2y/dx2
x=ln(1+t^2)
dx/dt=(1+t^2)'*(1/(1+t^2))=2t/(1+t^2)
y=arctant
dy/dt=1+t^2
dy/dx=(1+t^2)^2/2t=(e^x)^2/[2*√(e^x-1)] =e^(2x) /[2√(e^x-1)]
d^2y/dx^2=(e^2x)'/[2√(e^x-1)] +e^(2x)*(-1/2)*(e^x-1)'/[2*√(e^x-1)^3]
=e^(2x)/√(e^x-1) - e^(3x) /[4√(e^x-1)^3]

[(1+t^2)-t*2t]/(1+t^2)^2=(1-t^2)/(1+t^2)^2 y''=-2t/(1+t^2)^2 dy/dx=y'/x'=1/t d^2y/dx^2=(x'y''-x''y