x y=ln(xy)求隐函数y的导数dy dx
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方法一(微分法)d(y/x)=d(ln(xy))(xdy-ydx)/x²=1/xy*d(xy)即(xdy-ydx)/x²=(ydx+xdy)/xy∴dy/dx=(xy+y²
两边求导(y'x-y)/x^2=(y+xy')/xyxy+x^2y'=xyy'+y^2y'=(xy-y^2)/(xy+x^2)
首先z'(x)=x*(a-x-2*y)=0z'(y)=y(a-y-2*x)=0计算得到四组解(0,0)(a,0)(0,a)(a/3,a/3)1.(0,0)时,f''xx=0,f''xy=a,f''yy
z=xy+x/y对x的偏导数=y+1/y对y的偏导数=x-x/y^2
直接在等式中零,x=0,y=y(0),可得关于y(0)的方程解出y(0)即可.具体:e^0*y(0)+lny(0)/1=0即-y(0)=lny(0)作图y1=-x,y2=ln(x),两者的交点的横坐标
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
两边对x求导得y+xy'=(1+y')/(x+y)y(x+y)+x(x+y)y'=1+y'y'[x(x+y)-1]=1-y(x+y)y'=[1-y(x+y)]/[x(x+y)-1]dy=[1-y(x+
xdy=(y+xy)dxdy/y=((1+x)/x)dxln|y|=ln|x|+x+cy=±e^(ln|x|+x+c)其中c是常数再问:真还不理解我们是选择题:y=cxe^xy=c+x-x^2y=cs
设Y=y'降阶:Y'=(Y/x)ln(Y/x)这就是一个一阶齐次方程.设Y/x=u,所以Y=ux,Y'=u+x(du/dx),代回原方程,解得:lnu=C1x+1Y=xe^(C1x+1)所以y=[(C
y'=(y+xy')/(xy)xyy'-xy'=yy'=y/(xy-x)所以dy/dx=y'=y/(xy-x)
直接两边对x求导,得1/y*(-1/y2)*dy/dx=1/xy*(y+xdy/dx)下面会了吧
x=yln(xy)dx=d(yln(xy))=ln(xy)dy+(y/(xy))d(xy)=ln(xy)dy+(1/x)(ydx+xdy)=ln(xy)dy+(y/x)dx+dy合并同类项有(ln(x
这不是微分方程.你漏掉导数符号了或者漏掉微分符号d了.再问:没有,篇子上原题,一模一样。再答:你有没有看清楚,其中是不是有个y有个小小的一撇y'这真的不是微分方程,微分方程要含有导数或者偏导或者等价的
先求导等式两边同时对x求导得y+xy'+y'/y=0则y'=-y^2/(xy+1)当x=1,y=1时,y'=-1/2故切线方程为y-1=-1/2(x-1)即x+2y-3=0
[ln(xy)]'=[e^(x+y)]'(xy)'/(xy)=e^(x+y)*(x+y)'(y+xy')/(xy)=e^(x+y)*(1+y')y'=y[e^(x+y)-1]/[x(1-ye^(x+y
取定y=y0,lim(x--0)f(x,y0)=lim(x--0){ln(1+xy0)/x}=lim(x--0)(x*y0-x^2*y0^2+...)/x=lim(x--0)(y0-x*y0^2+..
x=yln(xy),等式两端对x求导,1=dy/dx+y[1/ln(xy)][y+x(dy/dx)]=dy/dx+y/ln(xy)+xdy/dx,整理得(dy/dx)(1+x)=1-y/ln(xy),
ln(xy)=e^x+ylnx+lny=e^x+y两边同时对x求导1/x+(1/y)(dy/dx)=e^x+dy/dxdy/dx=[(1/x)-e^x]/[1-(1/y)]=(y-xye^x)/(xy
两边求导得y'·e^y+(y+xy')/(xy)+e^(-x)=0
min是指f(x)g(x)h(x)三个函数中的最小值