求由方程y²-2xy 9=0所确定的隐函数y的导数

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'^

左右对x求导有y'/y=sec²(xy)(y+xy')整理有y'=y²/(cos(xy)-xy)所以dy=(y²/(cos(xy)-xy))dx

隐函数求导设z=x²y²-cos(xy)dy/dx=-(δz/δx)/(δz/δy)=-(2xy²+ysin(xy))/(2x²y+xsin(xy))=-y/x

-sin(xy)[ydx+xdy]=2xy^2*dx+x^2*2ydy-sin(xy)ydx-sin(xy)xdy=2xy^2*dx+2x^2*ydy-2x^2*ydy-sin(xy)xdy=2xy^

y^2/（x+y）=y^2-x^2y^2=（y^2-x^2）（x+y）两边同时求导得到：2yy’=(2yy’-2x)(x+y)+(y^2-x^2)(1+y’)2yy’=2yy’(x+y)-2x(x+y

红色圈出再问：那在试卷上怎么答呢再答：如果是大题目，直接写出这两个求导方程，像我这么叙述就行了，个人经验，仅供参考再问：能帮我再解以下另外那几个数学题吗再答：我尽力

两边对x求导,e^(2y)*2y'+3y+3xy'-2x=0,故dy/dx=y'=2x/[2e^(2y)+3x].

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

dy²-2d(xy)+0=02ydy-2(xdy+ydx)=02ydy-2xdy=2ydxdy/dx=y/(y-x)

方程两边同时对x求导得2yy'-3(y+xy')=0整理化简得y'=3y/(2y-3x)即dy/dx=3y/(2y-3x)

x-y+1/2siny=0两边对x求导得1-y'+1/2cosy*y'=0y'=2/(2-cosy)y''=dy'/dx=(dy'/dy)*(dy/dx)=[-2/(2-cosy)²]*si

siny+xe^y=0确定有隐函数：y=y(x)于是,同时在两边对x求导：(siny+xe^y)'=0'y'*cosy+e^y+xy'e^y=0y'*(cosy+xe^y)=-e^yy'=-e^y/(

对y^2－2xy＝7求微分,得2ydy-2(ydx+xdy)=0,∴（y-x)dy=ydx,∴dy/dx=y/(y-x).

x-y+1/2siny=0F(x,y)=y-x-1/2siny=0F,Fx,Fy在定义域的任意点都是连续的,F(0,0)=0Fy(x,y)>0f'(x)=-Fx(x,y)/Fy(x,y)=1/(1-1

xy+e^y=1e^y(0)=1y(0)=0xy'+y+e^yy'=00+y(0)+y'(0)=0y'(0)=0xy''+y'+y'+e^yy''+(y')^2e^y=00+2y'(0)+y''(0)

y^3z^2-x^2+xyz-5=0等式两边同时对x求导：∂z/∂x=（2x-yz）/（2zy^3+xy)等式两边同时对y求导：∂z/∂y=-(3y&#

设dy/dx=y'.求导,2yy'-2y-2xy'=0dy/dx=y'=y/(y-x)

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,e^y-e^0=0,则e^y=1,则y=0所以y'(

微分得xe^ydy+e^ydx+2ydy=0,解得dy/dx=-e^y/(xe^y+2y)

两边对x求导数,得y'*e^y+y+xy'=0,在原方程中令x=0可得y=1,因此,将x=0,y=1代入上式可得y'+1=0,即y'(0)=-1.再问：对x求导时y可以当成一个常数吗？为什么要用公式（