作业帮高数题设x=(t+1)e^t,y=t^2*e^t,求d^2y/dx^2
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高数题设x=(t+1)e^t,y=t^2*e^t,求d^2y/dx^2
高数题设x=(t+1)e^t,y=t^2*e^t,求d^2y/dx^2
高数题设x=(t+1)e^t,y=t^2*e^t,求d^2y/dx^2
参数方程求导：
d^2y/dx^2
=d[dy/dx] / dx
=d[(dy/dt) / (dx/dt)] / dx
=d[y'/x']/dt * dt/dx
=(y''x'-y'x'')/x'^2 * 1/x'
=(y''x'-y'x'')/x'^3
因此,
y=(t+1)e^t
y'=e^t+(t+1)e^t=(t+2)e^t
y''=e^t+(t+2)e^t=(t+3)e^t
x=t^2e^t
x'=2te^t+t^2e^t=(t^2+2t)e^t
x''=(2t+2)e^t+(t^2+2t)e^t=(t^2+4t+2)e^t
代入,得：
d^2y/dx^2
=(y''x'-y'x'')/x'^3
=[(t+3)e^t(t^2+2t)e^t-(t+2)e^t(t^2+4t+2)e^t] / [(t^2+2t)e^t]^3
=[(t+3)(t^2+2t)-(t+2)(t^2+4t+2)] / e^t(t^2+2t)^3
=[(t^2+3t)-(t^2+4t+2)] / e^t(t+2)^2t^3
=(-t-2) / e^t(t+2)^2t^3
=(-1) / e^t(t^4+2t^3)
即,
d^2y/dx^2=(-1) / e^t(t^4+2t^3)
当然做法不止一种,介绍一种计算上或许更简单的做法（分层）：
先算：
dy/dx
=dy/dt / dx/dt
=(t+2)e^t / (t^2+2t)e^t
=1/t
再算：
d^2y/dx^2
=d(dy/dx) / dx
=d(1/t)/dt / dx/dt
=-1/t^2 / (t^2+2t)e^t
=(-1) / e^t(t^4+2t^3)
有不懂欢迎追问
dy/dx=(dy/dt)/(dx/dt)
=(2te^t+t^2e^t)/(e^t+(t+1)e^t)
=(t^2+2t)/(t+2)
d^2y/dx^2=d(dy/dx)/dx
=(d(dy/dx)/dt)/(dx/dt)
=[((2t+2)(t+2)-(t^2+2t))/(t+2)^2]/(e^t+(t+1)e^t)
=1/e^t*(2t^2+6t+4-t^2-2t)/(t+2)^3
=1/e^t*(t^2+4t+4)/(t+2)^3
=1/(e^t(t+2))