www.whkt.net > 设z=ArCtAny/x,则z关于x的二阶偏导数是,求解答,急!

设z=ArCtAny/x,则z关于x的二阶偏导数是,求解答,急!

z=arctany/xδz/δy = 1/[1+(y/x)^2] * [1/x] = x/[x^2+y^2]δz/δy =δ(δz/δy)/δy = -x(2y)/[x^2+y^2]^2 = -2xy/[x^2+y^2]^2

对x求偏导,把y看成常数,对y求偏导,把x看成常数然后求导数.f'x= -arctany/x^2 f'xx=2arctany/x^3f'y= 1/[x(1+y^2 )] f''y=-2y/[x(1+y^2 )^2]

^解:设z=f(x,y) 则ez/ex=2xarctany/x-1/[1+(y/x)]-y^3/(x+y) =2xarctany/x-[xy/(x+y)]-y^3/(x+y) =(2xarctany/x)-y ∴ez/exey=[e(ez/ex)]/ey =2x(1/x)[1/(1+y/x)] =(x-y)/(x+y) 说明:这里用e代表偏导符号 祝你学习愉快!

偏导数的符号用a代替了哈 az/ax=1/tan(y/x)*sec^2(y/x)*(-y/x^2) =-y/(x^2sin(y/x)cos(y/x)) az/ay=1/tan(y/x)*sec^2(y/x)*1/x =1/(xsin(y/x)cos(y/x))

z=x^2arctan(y/x) dz=(z/x)dx+(z/y)dyz/x=2xarctan(y/x)+x[(1/(1+(yx))](-y/x))=2xarctan(y/x)-y/(1+(y/x))z/y=x[(1/(1+(y/x))](-1/x))=-x/(1+(y/x))

设f(x)=x^2+y^2+z^2-xf(y/x)=0 =x^2+y^2+z^2-xf(u)=0 u=y/xu/x=-y/x^2=-u/x, u/y=1/xf/x=2x-f(u)-x*f/u*u/x=2x-f(u)+f/u*uf/y=2y-x*f/u*u/y=2y-f/uf/z=2z对f(x)求全微分,得df=f/x*dx+f/y*dy+f/z*dz=0∴z/x=-(f/x)/(f/z)=-[2x-f(u)+f/u*u]/(2z) z/y=-(f/y)/(f/z)=-[2y-f/u]/(2z)

z=arctan(x/y) z/x = [1/(1+ (x/y)^容2)] /x ( x/y) = (1/y)[1/(1+ (x/y)^2)] = y/(x^2+y^2) z/y = [1/(1+ (x/y)^2)] /y ( x/y) = (-x/y^2)[1/(1+ (x/y)^2)] = -x/(x^2+y^2)

Zx=(2x+y)/[(x^2)+xy+(y^2)],Zy=(2y+x)/[(x^2)+xy+(y^2)]; Zxx=2/[(x^2)+xy+(y^2)]-[(2x+y)^2]/{[(x^2)+xy+(y^2)]^2}; Zxy=1/[(x^2)+xy+(y^2)]-[(2x+y)(2y+x)]/{[(x^2)+xy+(y^2)]^2}; Zyy=2/[(x^2)+xy+(y^2)]-[(2y+x)^2]/{[(x^2)+xy+(y^2)]^2}.

z/x=ln(x^2+y)+2x^2/(x^2+y),z/x=2x/(x^2+y)+4xy/(x^2+y)^2=2x(x^3+3y)/(x^2+y)^2,z/xy =1/(x^2+y)+(-2x^4)/(x^2+y)^2=(x^2+y-4x^3)/(x^2+y)^2,z/y =x/(x^2+y),z/y =-x/(x ^2+y)^2.若对x求偏导数,则把y当作常量,若对y求偏导数,则把x当作常量,用导数公式求偏导数,求混合偏导数,则与对x和y求偏导数的顺序无关.

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