设y=y(x)是由方程x^2 y^2-xy=4确定的隐函数
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xy+e^y=y+1(1)求d^2y/dx^2在x=0处的值:(1)两边分别对x求导:y+xy'+e^yy'=y'y/y'+x+e^y=1(2)(2)两边对x再求导一次:(y'y'-yy'')/y'^
d(y^2)/dx=d(y^2)/dy*dy/dx=2y*dy/dx这个复合函数求导法则正如ovtr0001仁兄所说那样,你可以翻翻课本这个……还要详细点呀?你有书么?你看书那里不懂可以提出来,我可能
对方程两边同时求导得,﹣﹙y+xy′﹚sin﹙xy﹚+e^y+﹙x+1﹚y′e^y=0令x=0则方程cos(xy)+(x+1)*e^y=2为1+e^y=2,得y=0,即切点坐标为﹙0,0﹚将﹙0,0﹚
lny+x/y=0等式两边求导:y'*1/y+1/y+x*y'(-1/y²)=0(1/y-x/y²)y'=-1/y∴y'=(-1/y)/(1/y-x/y²)=-y/(y-
两边对x求导2x+2y*dy/dx=0dy/dx=-x/y有不明白的追问再问:刚学不太明白,2x+2y*dy/dx=0里的dy/dx哪来的,是y'吗?再答:是的复合函数求导注意这里y是x的函数不妨换个
由隐函数微分法可得:-sin(x+y)(1+y′)+y′=0-sin(x+y)+[1-sin(x+y)]y′=0∴y′=sin(x+y)/[1-sin(x+y)].
设y=y(x)由方程ysinx=cos(x-y)所确定,则y'(0)=x=0时cos(-y)=cosy=0,故y=π/2+2kπ,k∈ZF(x,y)=ysinx-cos(x-y)=0dy/dx=-(&
这是一个复合函数求导,y=y(x)所以求e^y的导数首先对整体求导,再对y求导即为e^y*y'xy的导数为y+x*y'(根据求导规则)所以两边求导可得e^y*y'-y-x*y'=0
两边对x求导:2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问:求f'(x)```再答:y'就是f'(x)啊。。。。。
再答:隐函数高阶求导。再答:
对两边求导:[-sin(x+y)](1+dy/dx)+dy/dx=0-sin(x+y)-[sin(x+y)]dy/dx+dy/dx=0dy/dx=[sin(x+y)]/[1-sin(x+y)]
网上有很多高数课后习题答案,你可以下载一个参考~e^y-e^x=xy两边求导,得e^y*y'-e^x=y+xy'(e^y-x)y'=(e^x+y)所以y'=(e^x+y)/(e^y-x)x=0时,原式
分别对y求导,求左边为1+【e^(x+y)×(dx/dy+1)】右边为2×dx/dy推的dx/dy:自己算下,没得草稿纸.
答:x^2+y^2-xy=4两边对x求导:2x+2yy'-y-xy'=0(2y-x)y'=y-2xy'=(y-2x)/(2y-x)所以:dy=(y-2x)dx/(2y-x)
ln(x+y)=x·lny(1+y‘)/(x+y)=lny+x/y·y‘y+y·y‘=y(x+y)lny+x(x+y)·y‘y‘=【y(x+x)lny-y】/【y-x(x+y)】再问:лл����
首先du/dx=z+x*dz/dx而Z=Z(x,y)由方程x²z+2y²z²+y=0确定,对x求导得到2xz+x²*dz/dx+2y²*2z*dz/d
/>e^y+xy+e^x=0两边同时对x求导得:e^y·y'+y+xy'+e^x=0得y'=-(y+e^x)/(x+e^y)y''=-[(y'+e^x)(x+e^y)-(y+e^x)(1+e^y·y'
dy/dx=-fx/fy,你自己可以算吧
F(x,y)=x^2+y^2-ln(x+2y)Fx=2x-1/(x+2y)Fy=2y-2/(x+2y)F(x)=-Fx/Fy=-[2x(x+2y)-1]/[2y(x+2y)-2]