f(x,y)=(ysinx) (x^2 y^2) , x^2 y^2≠0求偏导数
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(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))
具体点的就是指数函数当然不一定了,a^(x+y)=a^x*a^y
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=-(&
参考答案:停车坐爱枫林晚,霜叶红于二月花.
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)
f'(f(f))*f'(f)*f'
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
假设(2xcosy+y^2*cosx)dx+(2ysinx-x^2*siny)dy某个函数u(x,y)的全微分du/dx=2xcosy+y^2*cosx.(1)du/dy=2ysinx-x^2*sin
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
f(x)f(y)=f((xy)/(1左边和右边有什么关系?就是左右相等啊,没别的意思这类的题目,一般都是将X或Y取一些特殊的值
假设:X=Y/XY=X/Y带入函数就是:F(y/x,x/y)=(y/x+x/y)/(y/x—x/y)=x²+y²)/(y²-x²)希望可以帮助你!
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