y=t-arctant求导
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![y=t-arctant求导](/uploads/image/f/911215-55-5.jpg?t=y%3Dt-arctant%E6%B1%82%E5%AF%BC)
X=arctantdx/dt=1/(1+t^2)y=ln(1+t2)dy/dt=2t/(1+t^2)dy/dx=(dy/dt)/(dx/dt)=2td2y/dx2=d(dy/dx)/dx=2dt/dx
x't=2ty't=1-1/(1+t^2)=t^2/(1+t^2)y'=dy/dx=y't/x't=t/[2(1+t^2)]d^y/dx^2=d(y')/dx=d(y')/dt/x't=1/2*[1+
直接对x求导算不出,所以先对t求导,再对x求导
dy/dx=(dy/dt)/(dx/dt)=[2t/(1+t^2)]/[1-1/(1+t^2)]=2/t
dy/dx=(dy/dt)/(dx/dt)显然dx/dt=1/(1+t²)给出的y是关于t的隐函数,可以不管这些,直接把y看成是t的函数,然后两边求导,得2dy/dt-(y²+2t
-(t^2+1)/(4t^3)dy/dt=1/(t*t+1)dx/dt=2t/(t*t+1)dy/dx=1/2td^2y/dx^2=[d(1/2t)/dt]*(t*t+1)/2t=-(t^2+1)/(
dy/dx=[1-1/(1+t²)]/[2t/(1+t²)]=t/2d²y/dx²=(1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1
dx/dt=1-2t/(1+t^2)=(1+t^2-2t)/(1+t^2)=(t-1)^2/(1+t^2)dy/dt=1/(1+t^2)y'=1/(t-1)^2dy'/dt=-2/(t-1)^3y"=
dy/dt=2t/(1+t²)dx/dt=1-[1/(1+t²)]=t²/(1+t²)dy/dx=(dy/dt)/(dx/dt)=2/t
书上给的公式也只有两阶导呀.
再问:e,那个dx/dt=2t/(1+t²)怎么来的??不应该是dx/d(1+t²)=1/(1+t²)然后怎么化的么。。。。。再答:这是复合函数的求导,先求出lnt的导数
楼主的补充问题,涉及到我们教学中长期普遍存在,却无人介意,更无人愿意更改的一个懒惰习惯,这是我们与英美教学的显著区别之处.点击放大,荧屏放大再放大:
答:x=ln(1+t²),x'(t)=2t/(1+t²)y=t-arctant,y'(t)=1-1/(1+t²)=t²/(1+t²)dy/dx=(dy
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/dxC中只含有t.因此,d^2y/dx^2=C/dt乘以dx/dt的倒数(dt/dx)
分别算出dx,dy,然后相除就行详见参考资料
dx/dt=2t/(1+t²)dy/dt=1/(1+t²)dy/dx=1/(2t)d(dx/dt)/dt=(2-4t²)/(1+t²)²d(dy/dt
∵dx=2tdt/√(1+t²)dy=dt/(1+t²)∴dy/dx=[2tdt/√(1+t²)]/[dt/(1+t²)]=2t√(1+t²).
x=tany+ln(cosy^2),dy/dx=(dx/dy)^-1=(tany-1)^-2,y"=d(dy/dx)/dy*dy/dx=-2secy^2/(tany-1)^5