y=f(x)由方程y³ xy² x²y 6确定,求f(x)极值
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/17 21:50:36
![y=f(x)由方程y³ xy² x²y 6确定,求f(x)极值](/uploads/image/f/910086-6-6.jpg?t=y%3Df%28x%29%E7%94%B1%E6%96%B9%E7%A8%8By%C2%B3+xy%C2%B2+x%C2%B2y+6%E7%A1%AE%E5%AE%9A%2C%E6%B1%82f%28x%29%E6%9E%81%E5%80%BC)
方程两边同时求x对y的导:y+xdy/dx+1/x+2ydy/dx=0,dy/dx=-(y+1/x)/(x+2y),dy=-(y+1/x)dx/(x+2y)
两边对x求导,(y+xy')e^xy=2+3y'代入(0,1)1=2+3y',y'=-1/3(y-1)=-x/3整理,得x+3y-3=0
两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)
直接两边对X求导,注意Y是X的函数.所以得:y+xy'=e^(x+y)*(1+y'),化简,代入原方程得:y+xy'=xy(1+y'),然后对得到的式子在此求导,得:y'+y'+xy''=(y+xy'
两边同时对X求导y+xy`=e^x+y`y`=(e^x-y)/(x-1)
这个题目要利用隐函数的求导法则.则sin(x^2+y)=xy(两边同时求导,还要结合复合函数的求导法则)cos(x^2+y)*(2x+y′)=y+xy′2xcos(x^2+y)-y=xy′-y′cos
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+
在xy+e^xy+y=e两边同时进行取微分,ydx+xdy+e^xy*(ydx+xdy)+dy=0然后求出dy/dx求出来后,在dy/dx等式两边两边同时求导,求导的过程中会有dy/dx,带入第一步求
y是x的函数,对x求导则e^(x²)*(x²)'-2y*y'=x'*y+x*y'2xe^(x²)-2y*y'=y+x*y'y'=[2xe^(x²)-y]/(x+
证明令x=x/y,y=y∵f(xy)=f(x)+f(y)∴f(x/y*y)=f(x/y)+f(y)f(x)=f(x/y)+f(y)∴f(x/y)=f(x)-f(y)
e^y-e^x+xy=0e^y*y’-e^x+y+xy'=0y'=(e^x-y)/(e^y+x)
xy+y^2-2x=0y+xy'+2yy'-2=0(x+2y)y'=2-yy'=(2-y)/(x+2y)dy/dx=(2-y)/(x+2y)
两端对x求导数(把y看作x的函数),则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]
Fx=e^x-y^2Fy=cosy-2xydy/dx=-Fx/Fy=(y^2-e^x)/(cosy-2xy)
(cos(x+y)-y)\(x-cos(x+y))
令y=kxx*x+kx*x=k*x+k*k*x*x(1-k*k+k)x^2-kx=0x((1-k*k+k)x-k)=0由上式得X=0或(1-k*k+k)x-k=0解得:k=(x-1+(或-)√((1-
两边对x求导xy^2+sinx=e^yy^2+2xyy'+cosx=e^y*y'y'(e^y-2xy)=y^2+cosxy'=(y^2+cosx)/(e^y-2xy)
两边对x求导:3x^2-3y^2-6xyy'+6y^2y'=0得y'=(y^2-x^2)/[2(y^2-xy)]=(y+x)/(2y)令y'=0,得y+x=0,将y=-x代入原方程:x^3-3x^3-