ysinx-cos(x-y)=0 求隐函数

（一题）从这步d(ysinx)-dcos(x-y)=0到这步sinxdy+ycosxdx+sin(x-y)(dx-dy)=0不懂是么?ysinx是两个数相乘,对它d(ysinx)时就得用公式d(UV)

ysinx=cos(xy)两边分别求导y'sinx+ycosx=-sin(xy)(y+xy')y'=-y(sin(xy)+cosx)/(sinx+xsin(xy))

两边关于x求一阶导y'*e^(x+y)-y'sinx-ycosx=0y'=ycosx/(e^(x+y)-sinx)

再答：是（x+y）^2还是x+y^2再问：是前者再问：第一道题你算错了吧。再答：为啥。。。。再问：再问：这个是答案。再答：第二个你把分子分母倒一下。。。。我看看。。？再问：？？再问：再问：第二道题再答

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=-(&

COS（X+Y）COS（X-Y）=(COSX*COSY-SINX*SINY)(COSX*COSY+SINX*SINY)=(COSX*COSY)^2-(SINX*SINY)^2=COS^2X(1-SIN

把函数y=cos（x+43π）的图象向右平移φ个单位，可得函数y=cos（x-φ+43π）的图象；再根据所得图象正好关于y轴对称，可得-φ+43π=kπ，k∈z，即φ=-kπ+4π3，故φ的最小正值为

参考答案:停车坐爱枫林晚,霜叶红于二月花.

symsxyeq=cos(x*y)*cos(x*(1-y))-0.5*x*sin(x*y)*sin(x*(1-y))-1;ezplot(eq)

两边对x求导：dy/dxsinx+ycosx-sin(x-y)(1-dy/dx)=0,将x=π/2带入已知方程得到y,再把x、y带入上式求得结果再问：x=π/2带入已知方程得到y。。。我算不出这个y

这就是应用隐函数的求导.将x=0代入方程,得lny^2=0,得y=±1两边对x求导,得:（2x+2yy')/(x^2+y^2)=y'sinx+ycosx+1代入x=0,y=1到上式,得2y'=2,得y

ysinx-cos(x+y)=0,两边对x求导,得y'sinx+ycosx+(1+y')sin(x+y)=0,解得y'=-[ycosx+sin(x+y)]/[sinx+sin(x+y)]dy/dx=y

-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k

Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2

应用复合函数求导方法,y′sinx+ycosx+（1+y′）sin（x+y）=0,（sinx+sin（x+y））y′+ycosx+sin（x+y）=0,y′=-（ycosx+sin（x+y））/（si

两边对x求导y'*sinx+ycosx-[-sin(x+y)*(1+y')]=0y'(sinx+sin(x+y))=y(1-cosx)y'=[1-cosx]/[sinx+sin(x+y)]0/0所以需

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两边求导：y'sinx+ycosx+sin(x+y)*(1+y')=0令x=0,y=π/2：π/2+1+y'=0y'=-(π/2+1)dy=-(π/2+1)dx

1.两边求导得：y'=-sin(x-y)(1-y')解得y'=sin(x-y)/[sin(x-y)-1]2.y'=-e^-xy''=e^-xy'"=-e^-x3.y'"=(e^2x)'"(sinx)+