x'=t/（1+t^2)
y'=1/（1+t^2)
x''= [（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')/(x')^3
=[-2t^2/（1+t^2)^3-(1-t^2)/（1+t^2)^3]/[t/（1+t^2)]^3
=(-t^2-1)/t^3
设{x=ln√(1+t^2),y=arctant, 求 dy\/dx及d^2·y\/d·x^2 有详细过程最...
y''=-2t\/（1+t^2)^2 dy\/dx=y'\/x'=1\/t d^2y\/dx^2=(x'y''-x''y')\/(x')^3 =[-2t^2\/（1+t^2)^3-(1-t^2)\/（1+t^2)^3]\/[t\/（1+t^2)]^3 =(-t^2-1)\/t^3
求由参数方程x=ln√(1+t^2) y=arctant所确定的函数的导数求d^2y\/...
答：x=ln√(1+t^2)，dx\/dt=[1\/√(1+t^2)]*(1\/2)*2t\/√(1+t^2)=t\/(1+t^2)y=arctant，dy\/dt=1\/(1+t^2)所以：dy\/dx=1\/t y''=d²y\/dx²=d(dy\/dx)\/dx=[d(1\/t)\/dt]\/(dx\/dt)=(-1\/t^2)\/[t\/(1+t^2)]=-(1+t^2)\/t^3 所以：dy\/dx=1...
x=ln√(1+t^2),y=arctant。求d2y\/dx2 要过程哦
解答 如图
x=ln(1+t^2),y=t-arctant 求d^2y\/dx^2的导数,详细过程啊!
先分别求出dx\/dt和dy\/dt，假设A=dx\/dt ,B=dy\/dt 然后用B\/A 得出dy\/dx 设C=B\/A=dy\/dx C中只含有t。因此， d^2y\/dx^2=C\/dt乘以dx\/dt的倒数（dt\/dx）=C\/dx=(dy\/dx)\/dx PS:式子A，B，C是简单的求导计算，这里就不计算了 ...
设参数方程为x=ln(1 t^2) y=arctant,求yd^2\/dx^2
猜x=ln(1+t^2),y=arctant,则 dx\/dt=2t\/(1+t^2),dy\/dt=1\/(1+t^2),∴dy\/dx=1\/(2t),于是d^2y\/dx^2=d(dy\/dx)\/dt*dt\/dx =-1\/(2t^2)*(1+t^2)\/(2t)=-(1+t^2)\/(4t^3).
设参数函数x=ln(1+t^2),y=t-arctant. 求(d^2y)\/(dx^2).
先分别求出dx\/dt和dy\/dt，假设a=dx\/dt ,b=dy\/dt 然后用b\/a 得出dy\/dx 设c=b\/a=dy\/dx c中只含有t。因此，d^2y\/dx^2=c\/dt乘以dx\/dt的倒数（dt\/dx）=c\/dx=(dy\/dx)\/dx ps:式子a，b，c是简单的求导计算，这里就不计算了 ...
方程组 x=ln√1+t^2 y=arctant 求 dy\/dx 详细步骤及解析 希望可以有语 ...
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求导数的相关题 x=ln(1+t^2) y=arctant 求d^2 y\/dx^2=
-(t^2+1)\/(4t^3)dy\/dt=1\/(t*t+1)dx\/dt=2t\/(t*t+1)dy\/dx=1\/2t d^2 y\/dx^2=[d(1\/2t)\/dt]*(t*t+1)\/2t=-(t^2+1)\/(4t^3)
求参数方程x=ln(1+t∧2),y=t-arctant,所确定函数的三阶导数。详细过程最...
x't=2t\/(1+t^2)y't=1-1\/(1+t^2)=t^2\/(1+t^2)y'=dy\/dx=y't\/x't=t\/2 y"=d(y')\/dx=d(y')\/dt \/(dx\/dt)=(1\/2)\/[2t\/(1+t^2)]=(1+t^2)\/(4t)=1\/4*[1\/t+t]y"'=d(y")\/dx=d(y")\/dt\/(dx\/dt)=1\/4*[-1\/t^2+1]*(1+t^2)\/(2t)=(t...
x=arctant y=ln(1+t^2) 求 d^2\/dx^2
x=arctant dx\/dt = 1\/(1+t^2)y=ln(1+t^2)dy\/dt = 2t\/(1+t^2)dy\/dx = (dy\/dt)\/( dx\/dt) = 2t d^2y\/dx^2 =d\/dx ( dy\/dx)=d\/dt ( dy\/dx) \/ [ dx\/dt]=(1+t^2) .d\/dt ( dy\/dx)=(1+t^2) .d\/dt ( 2t)=2\/(1+t^2)
