z+cos+x+y+对x求偏导
耿疫迹1111已知复数Z=x+yi(x,y属于R)满足|Z - 1|=1,求复数Z的摸取值范围 -
武雍凯19847976793 ______ |z-1|=√[(x-1)^2+y^2]=1 所以(x-1)^2+y^2=1 设x=sint+1 y=cost 则 |z|=√[x^2+y^2]=√[(sint+1)^2+cos^2t]=√(sin^2t+2sint+1+cos^2t)=√(2sint+2) 因为-1所以0即0
耿疫迹1111已知x,y,z都是锐角,且cosx+cosy+cosz=1+4sinx/2*siny/2*sinz/2,求证:x+y+z=π -
武雍凯19847976793 ______ cosx+cosy+cosz=1+4sin(x/2)sin(y/2)sin(z/2)cosx+cosy=1-cosz+4sin(x/2)sin(y/2)sin(z/2)2cos[...
耿疫迹1111求z=sin(x+y)e^x*y在点(1, - 1)处的偏导数 -
武雍凯19847976793 ______ zx=cos(x+y)*ye^xy zx(1,-1)=-e^(-1) zy=cos(x+y)*xe^xy zy(1,-1)=e^(-1)
耿疫迹1111二元函数z=cos3xy+ln(1+x+y)的全微分dz= -
武雍凯19847976793 ______ dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy
耿疫迹1111u=x(z+y) z=sin(x+z) 求二阶偏导数σ2u/σxσy -
武雍凯19847976793 ______ z是x 的函数, 于是u 是x 和y 的函数. z=sin(x+z) => z ' = cos(x+z) ( 1+ z ' ) => dz/dx = cos(x+z) / [ 1- cos(x+z) ] u= F(x,y,z) = x(z+y), δu/δx = δF/δx + δF/δz * dz/dx = z+y + x * cos(x+z) / [ 1- cos(x+z) ] δ²u/δxδy = δ (δu/δx) /δy = 1
耿疫迹1111已知方程sin(x+y)+sin(y+z)=1确定了隐函数z=f(x,y),求∂z∂x,∂z∂y及∂2z∂x∂y. -
武雍凯19847976793 ______[答案] 对方程sin(x+y)+sin(y+z)=1两边关于x,y求偏导,得 cos(x+y)+cos(y+z) ∂z ∂x=0, cos(x+y)+cos(y+z)(1+ ∂z ∂y)=0, ∴ ∂z ∂x=- cos(x+y) cos(y+z), ∂z ∂y=- cos(x+y) cos(y+z)-1. 又对方程cos(x+y)+cos(y+z)(1+ ∂z ∂y)=0两边关于x求导,得 -sin(x+y)-sin(y+...
耿疫迹1111求二阶偏导数z=xsin(x+y) -
武雍凯19847976793 ______[答案] z=xsin(x+y) ∂z/∂x=sin(x+y)+xcos(x+y) ∂z/∂y=xcos(x+y) ∂²z/∂x²=cos(x+y)+cos(x+y)-xsin(x+y) =2cos(x+y)-xsin(x+y) ∂²z/∂y²=-xsin(x+y) ∂²z/∂x∂y=cos(x+y)-xsin(x+y)
耿疫迹1111u=x(z+y) z=sin(x+y) 求二阶偏导数σ2u/σxσy -
武雍凯19847976793 ______[答案] σu/σx=(z+y)+x(σz/σx+0)=z+y+xcos(x+y) σ2u/σxσy=σz/σy+1-xsin(x+y)=cos(x+y)+1-xsin(x+y)
耿疫迹11111 已知x.y,z为锐角,cos²x+cos²y+cos²z=1,求证:3π/4
耿疫迹1111将sinx+siny+sinz - sin(x+y+z)化为积的形式 -
武雍凯19847976793 ______ 解:原式=sinx+siny+sinz-sin(x+y+z) =4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2] =2sin[(x+y)/2]cos[(x-y)/2]+sinz-sin(x+y)cosz-sinzcos(x+y) =2sin[(x+y)/2]cos[(x-y)/2]+sinz[1-cos(x+y)]-sin(x+y)cosz =2sin[(x+y)/2]*{cos[(x-y)/2]+sinzsin[(x+y)/2]-cos[(x+y)/2]cosz} =2...