xy=e^x y的隐函数y的导数
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直接求导(xy^2)=y^2+2xy*y'(e^xy)'=(xy)'*xy*e^xy=(y+x*y')*xy*e^xy然后带进去求y'就是dy/dx
隐函数求导,两边同时求导,此题是对X求导!两边同时求导:y+xy'=e^x-y'y'=(e^x-y)/(x+1)由XY=e^X-y解出yy=e^x/x+1,带入上式y'=(e^x-y)/(x+1)=[
方程两边对x求导:e^y×y'=y+xy'得y'=y/(e^y-x)
xy=e^x+yxy'+y=e^x+y'y'(1-x)=y-e^xy'=(y-e^x)/(1-x)
e^y+xy-e=0d(e^y)+d(xy)-d(e)=0e^ydy+xdy+ydx=0(e^y+x)dy=-ydxdy/dx=-y/(e^y+x)
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
你明白复合函数吗?你的求导是对x求导,然后y是关于x的函数,y可以x表示,所以e^y=e^y*(y'),因为是对x求导,所以要加上dy/dx..类比于e^x对x求导,是e^x*(dx/dx)=e^x
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.
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?
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'
[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
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))
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
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+
xy=e^x-e^y两边求导得:y+xy'=e^x-y'*e^y解得:y'=(e^x-y)/(e^y+x)
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)