f+x+xy+对x求偏导

阚建彪3017设u=x²f(x+y,x - y),f(u,v)的一阶偏导数连续,求∂u/∂x, ∂u/∂y. -
戚饶俭18830641943 ______ u=x²f(x+y,x-y) 那么对x 求偏导得到 ∂u/∂x=(x²)' *f +x² *f1' *∂(x+y)/∂x +x² *f2' *∂(x-y)/∂x 而显然∂(x+y)/∂x=∂(x-y)/∂x=1 所以得到 ∂u/∂x=2x *f + x² *f1' +x² *f2' 同理对y 求偏导得到 ∂u/∂y=x² *f1' *∂(x+y)/∂y +x² *f2' *∂(x-y)/∂y 而∂(x+y)/∂y=1,∂(x-y)/∂y= -1 故解得 ∂u/∂y=x² *(f1' -f2')

阚建彪3017设f(x)在0到正无穷有定义,且f'(X)=a(a不=0),对任意x,y(属于0到正无穷)有f(xy)=f(x)+f(y),求f(x) -
戚饶俭18830641943 ______ 解:f'(X)=a得f(x)=ax+b 由f(xy)=f(x)+f(y)得 axy+b=(ax+b)(ay+b)=a^2*xy+ab(x+y)+b^2对任意的x,y恒成立.于是a=a^2,考虑到a≠0,故a=1.则xy+b=xy+b(x+y)+b^2恒成立.于是必须b=0.故f(x)=x

阚建彪3017求助 一个问题
戚饶俭18830641943 ______ F(a, b)其中a = x + y,而b = xy,可知x, y是自变量,f为函数. F对x求导,发现a和b中均含有x,于是,需要F对a,b分别求偏导数,然后a、b再对x求偏导数 即dF/dx = dF/d(x+y) * d(x+y)/dx + dF/d(xy) * d(xy)/dx 而dF/dx用F1表示,即对第一个变量求偏导. dF/d(xy)用F2表示. 求出1阶偏导后,比如F1 + y*F2 把F1看作新的求导对象,从而得到了F11 + y*F12这样的二阶偏导....

阚建彪3017设z(x,y)是由f(x+y,y+z)=0构成的函数 求dz -
戚饶俭18830641943 ______ f(x+y,y+z)=0分别对x、y求偏导 f1*dx+f2*(z'x)dx=0 f1*dy+f2*dy+f2*(z'y)dy=0 于是(z'x)dx=-dx*f1/f2(z'y)dy=-(f1*dy+f2*dy)/f2 所以dz=(z'x)dx+(z'y)dy=-(f1*dx+f1*dy+f2*dy)/f2

阚建彪3017f(x+y,xy)=x^2 - xy+y^2,则f'x(1,1)= -
戚饶俭18830641943 ______ 由 f(x+y,xy)=x^2-xy+y^2=(x+y)^2-3xy 原函数 f(x,y)=x^2-3y 对x的偏导 f'x(x,y)=2x f'x(1,1)=2 希望对你有点帮助!

阚建彪3017求函数f(x,y)=x2+xy+y2 - 6x - 3y的极致 -
戚饶俭18830641943 ______ 解:分别对x,y求偏导数得:f'(x)=2x+y-6 f'(y)=2y+x-3 令两者都为0,解得驻点为:(3,0) 又分别对其求二阶偏导数:f''(x)=2 =A f''(y)=2 =C 用f'(x)再对y求偏导数得:f''(x,y)=1 =B 由极值的判别式可得:[f''(y)]^2 - [f''(x)].[f''(x,y)]=B^2 - A.C =1-4=-3<0 又因为:A=2>0,故在(3,0)这点取得极小值:f(x,y)极小=f(3,0) =9-18=-9

阚建彪3017设z=(x,y)是由方程f(x+y,y+z)=0所确定的隐函数,则dz=? -
戚饶俭18830641943 ______ f(x+y,y+z)=0 两边同时取微分,得 f1 d(x+ f1(dx+dy)+f2 (dy+dz)=0 f1dx+(f1+f2)dy=-f2dz 所以 dz=-f1/f2 dx-(f1+f2)/f2 dy

阚建彪3017设u=f(x+y,x - y)具有二阶连续偏导数,求u对X的连续二次偏导数 -
戚饶俭18830641943 ______[答案] 对 u = f(x+y,x-y) 关于 x 求偏导数,得 Du/Dx = (D/Dx)f(x+y,x-y) = f1(x+y,x-y)*(D/Dx)(x+y)+f2(x+y,x-y)*(D/Dx)(x-y) = f1(x+y,x-y)*1+f2(x+y,x-y)*1 = f1(x+y,x-y)+f2(x+y,x-y), D(Du/Dx)/Dx = (D/Dx)[f1(x+y,x-y)+f2(x+y,x-y)] = [f11(x+y,x-y)+f12(x+y,x-y)]+[f...

阚建彪3017xy+yz+zx=1 确定z=f(x,y),隐函数求偏导的问题、 设Fx=xy+yz+zx - 1;那Fx=y+y(f1)+x(f1)+z 有错吗 -
戚饶俭18830641943 ______[答案] 错的是z对x的偏导,不是Fx z对x的偏导 xy+yz+zx=1 得到y+yfx'+z+xfx'=0 也就是Zx=y+y(f1)+x(f1)+z

阚建彪3017x^2+y^2=e^(arctan(y/x)),求dy/dx令F(x,y)=x^2+y^2 - e^arctan(y/x)=0对x求偏导 Fx = 2x - e^arctan(y/x) * 1/[1+(y/x)^2] * [ - y/(x^2)]=2x + y/(x^2+y^2) * e^arctan(y/x)然后e... -
戚饶俭18830641943 ______[答案] x^2+y^2-e^(arctan(y/x))=0 2x+2y*y'-(arctan(y/x))'e^(arctan(y/x))=0 2x+2y*y'-1/(1+y^2/x^2)*(y'x-y)/x^2 *e^(arctan(y/x))=0 2x+2y*y'-(y'x-y)/(x^2+y^2)*e^(arctan(y/x))=0 2x+2y*y'-y'x/(x^2+y^2)*e^(arctan(y/x))+y/(x^2+y^2)*e^(arctan(y/x))=0 y'=(2x++y/(x^2+y^2)*e^(...