求微分y方程x(y^2-1)dx y(x^2-1)dy=0

隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^

第一题,这是个隐函数,两边对x求导得：2y'-1=(1-y')*ln(x-y)+(x-y)*(1-y')/(x-y)=(1-y')*ln(x-y)+(1-y')所以[3+ln(x-y)]y'=ln(x

第二个你说的不对.这实际上是复合函数的求导.dtan^2(1-x)=2tan(1-x)*dtan(1-x)=2tan(1-x)*sec^2(1-x)d(1-x)=-2tan(1-x)sec^2(1-x

(1)y=3x^2-ln1/x=3x^2+lnxdy=6xdx+(1/x)dx=(6x+1/x)dx(2)y=e^(-x)cosxdy=-e^(-x)cosxdx-e^(-x)sinxdx=-e^(-

两边对x求导2xy+x^2y'-(1+y^2)^(1/2)*y'=0前面两项是对于原方程的第一项运用积法则+链式法则得来的整理可得y'=2xy/[(1+y^2)^(1/2)-x^2]

-((2x)/(1-x^2))dx;(-E^-x-Sin[3+x])dx;2Cos[2x]dx

dx/dt=-e^(-t)sint+e^(-t)cost=e^(-t)(cost-sint)dy/dt=e^tcost+e^t(-sint)=e^t(cost-sint)dy/dx=(dy/dt)/(

1、y=2x/(x+1),求y'（0）y'=2/(x+1)-2x/(x+1)^2=2/(x+1)^2y'（0）=22、y=ln(x^2+3),求dyy'=1/(x^2+3)*2x=2x/(x^2+3)

F(x,y)=x+lny-y=0dF(x,y)=0=(∂F(x,y)dx/∂x)+(∂F(x,y)dy/∂y)dy/dx=-(∂F(x,y)

如果对x求导,则ln|x|=yln|y|,1/x=y'/y+yy'/y=y'/y+y',.对数求导法.如果对y求导,则ln|x|=yln|y|,x'/x=ln|y|+y/y,x'=y^y(1+ln|y

dy＝｛1/√［1－（x^2－1）］｝d［√（x^2－1）］＝［1/√（2－x^2）］｛1/［2√（x^2－1）］d（x^2－1）＝｛x/√［（2－x^2）（x^2－1）］｝dx

1.d(cosx)^2=2cosx(-sinx)dx=-sin2xdx2.dsin(x²-1)=cos(x²-1)d(x²-1)=cos(x²-1)×2xdx=

y=arcsin√(1-x^2)y'=-x/(|x|√(1-x^2))∴dy=-xdx/(|x|√(1-x^2))当x>0dy=-dx/√(1-x^2)当x

y=(1+1/√x)(1-√x)=[(1+√x)/√x](1-√x)=(1-x)/√x=1/√x-√xdy= -1/2*x^(-3/2)dx-1/2*x^(-1/2)dx=-dx/(2x√x

y=[ln(1-x)^2]^2y'=2[ln(1-x)^2]*[ln(1-x)^2]'=2[ln(1-x)^2]*[2ln(1-x)]'=2[ln(1-x)^2]*2*1/(1-x)=4*[ln(1-

复合函数求导法则：y=u,u=v,v=f(x)=>dy/dx=dy/du*du/dv*dv/dx

因为y=y(x),所以y是x的函数!这道题考虑的主要是隐函数求导的问题.因为：y-2x=(x-y)ln(x-y)所以,将上式两边关于x求导后得：→y'-2=(1-y')ln(x-y)+(x-y)[1/

y'=2e^2xcos(e^2x)把y看成复合函数sint,t=e^m,m=2x.复合函数求导,等于三个分别求导的积

-sinx-2x