x-y (1 2)siny=0 求二阶导数
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z对x的偏导=cosx+cos(x+y)=0时,cosx=-cos(x+y)=cos(pi-x-y),所以x=pi-x-y.同理z对y的偏导=0时,有y=pi-x-y.所以x=y=pi/3.此时z=3
再问:大哥,你题目看错了。。。再答:哪里有错?再问:第一条等式就错了。。是sin(x+y)=sinx+siny。后面是cos(x+y)·(1+y')=cosx+cosy·y'?再答:OK,那我改下
两边求导:cos(x+y)*(1+y')=cosx+cosy*y'y'=(cosx-cos(x+y))/(cos(x+y)-cosy)e^x+1=e^y*y'+y'y'=(e^x+1)/(e^y+1)
两边对x求两次导数:1-y'+1/2cosyy'=0;==>y'=1/(1-cosy/2)0-y''+1/2(y'(-siny)+cosyy'')=0==>y''=y'siny/(cosy-2)再将y
再问:还有二阶导
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
1+y'=cosy*y'y'=1/(cosy-1)dy/dx=1/(cosy-1)
应该是2x^(2y)y'-2e^(2X)-cos(y)y'=0再问:2x^(2y)y'里2y的项是怎么来的呢?再答:好像求错了,是x^(2y-1)+2x^(2y)ln(x)y'-2e^(2X)-cos
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/(
变为dx/dy=-x+siny公式:对于y'=P(x)y+Q(x),通解为y=(∫{Q(x)e^[-∫P(x)dx]}dx+C)e^[∫P(x)dx]对于dx/dy=-x+siny,P(y)=-1,Q
对x求导得到1-y'+0.5cosy*y'=0所以y'=1/(1-0.5cosy)而继续求导得到y"=-1/(1-0.5cosy)^2*(0.5siny)*y'再代入y'即y"=-1/(1-0.5co
两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)
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
两边关于x求导,注意y是x的函数y'cosy=[1/(x+y)]*(1+y').①解得y'=1/(x+y)÷[cosy-1/(x+y)].②对①两边关于x求导可得y''cosy-(y')²s
cos(x+y)cosy+sin(x+y)siny=cos((x+y)-y)=cosx=4/5sinx=正负3/5tanx=正负3/4
x*e^y+siny=0e^y+x*e^y*y'+cosy*y'=0=>y'=-e^y/[xe^y+cosy]再问:你好!我数学太烂。。能不能补充一下完整的答案。。。再答:x*e^y+siny=0两边
sinx+siny=-sinzcosx+cosy=-cosz平方相加sin²x+cos²x+sin²y+cos²y+2(cosx+cosy+sinxsiny)=
这是隐函数的求导cosy*y'+3e^3x-6x^2y^2-4x^3*y*y'=0dy/dx=y'=(6x^2y^2-3e^3x)/(cosy-4x^3y)
两边对x求导有1-y'+y'cosy=0所以y'=1/(cosy-1)