xy=e^x y的导数
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这种题可以直接全微分,即e^xdx+xdy+ydx=0所以dy/dx=(e^x+y)/-x
对x求导为y*e^(xy)对y求导为x*e^(xy)对x,y求偏导为e^(xy)+xy*e^(xy)
xy=e^x+yxy'+y=e^x+y'y'(1-x)=y-e^xy'=(y-e^x)/(1-x)
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
先移项:e=e^y+xy,再两边对x求导:0=e^y*y'+y+x*y',解得:dy/dx=y'=-y/(e^y+x)
该题为隐函数求导.xy+e^(xy)=1则y+xy'+e^(xy)(y+xy')=0解得:y'=-y/x解答完毕.
即对x求导嘛.即(a*b)'=a'*b+a*b',上式a=x,b=e^-xy,x'=1,e^-xy=-y*e^-xy,整理就得结果啦
先对X求导y+xy'-e^x+e^yy'=0y'=(e^x-y)/(x+e^y)再问:主要是e^y我不懂,答案是对的,老师。还有y'=0是为什么?
答案是1/e当x=1,y=ln(0*1+e)=lne=1所以(0,1)在曲线上.y=ln(xy+e)y'=1/(xy+e)*(y+x*y')y'=y/(xy+e)+x/(xy+e)*y'y'*[1-x
构造函数,F(X,Y)=xy-e^(xy)则dy/dx=-Fx/Fy=-[y-e(xy)*y]/[x-e^(xy)*x]
两边求导:e^(xy)*(xy)'-(xy)'=0e^(xy)*(y+xy')-(y+xy')=0ye^(xy)+xe^(xy)*y'=y+xy'x(e^(xy)-1)y'=y(1-e^(xy))y'
隐函数求导,就是先左右一起求微分,加个d,然后写出多少dx+多少dy=0,移项变成dy/dx=多少的形式就好了
xy=e^x-e^yd(xy)=d(e^x-e^y)xdy+ydx=e^xdx-e^ydy(x+e^y)dy=(e^x-y)dx则由dy/dx=(e^x-y)/(e^y+x)
边对x求导有y+xy'=e^(x+y)*(1+y')解得dy/dx=y'=(e^(x+y)-y)/(x-e^(x+y))
解两边求导y‘cosy+e^x-y^2-2xyy'=0即y’(cosy-2xy)=y^2-e^xy'=(y^2-e^x)/(cosy-2xy)或者F(x,y)=siny+e^x-xy^2=0Fx=e^
两边分别求x的导数得:e^x+(y+xy')=0,即y'=-(e^x+y)/x,即:dy/dx=-(e^x+y)/x
(xy)'=(e^(x+y)'y+xy'=e^(x+y)*(1+y')y'=[e^(x+y)-y]/[1-e^(x+y)]
对x求导y+x*y'=e^(x+y)*(1+y')y+x*y'=e^(x+y)+e^(x+y)*y'所以dy/dx=[e^(x+y)-y]/[x-e^(x+y)]
所谓隐函数、只是说它的解析式其本质也是Y是X的函数,X为自变量第一道题中的y+x(dy/dx)都是xy对x求导的结果这是两个函数相乘求导(uv)'=u'v+uv'而e导数就为0第二道题也是一样-2y+
e^(x+y)=xy两边对x求导:e^(x+y)*(1+y')=y+xy'解得:y'=[y-e^(x+y)]/(e^(x+y)-x]=(y-xy)/(xy-x)