x=(t^2-2)\(t+1) y=(2t)\(t+1) 求dy\dxx=(t^2-2)\(t+1)y=(2t)\(t+1)求dy_数学解答

x=(t^2-2)\(t+1) y=(2t)\(t+1) 求dy\dx
x=(t^2-2)\(t+1)
y=(2t)\(t+1)
求dy\dx

人气：286 ℃　时间：2019-09-05 16:34:29

解答

dy/dt=((2t)'(t+1)-(2t)(t+1)')/(t+1)^2=(2(t+1)-2t)/(t+1)^2=2/(t+1)^2
dx/dt=((t^2-2)'(t+1)-(t^2-2)(t+1)')/(t+1)^2=(2t(t+1)-(t^2-2))/(t+1)^2=(t^2+2t+2)/(t+1)^2
所以dy/dx=(dy/dt)/(dx/dt)=2/(t^2+2t+2)