Z=xsin(xy)一阶偏导数-搜答案-拍题作业网

Z=f'x(x,y)=xy*[x^(xy-1)]*yZ=f'y(x,y)=xy*[x^(xy-1)]*x再问：答案是Z=f'x(x,y)=yx^xy(lnx+1),Z=f'y(x,y)=x^(xy+1

根据一阶全微分形式不变得dz=d(xf(x^y,e^xy)=f(x^y,e^xy)dx+xd(f(x^y,e^xy))=f(x^y,e^xy)dx+x[f1'd(x^y)+f2'(de^xy)]=f(

∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/&#

令u=x^2+y^2,v=xy得∂z/∂x=(∂f/∂u)(∂u/∂x)+(∂f/∂v)(∂

∂Z/∂x=y*cos(xy)-2cos(xy)*sin(xy)*y=y*cos(xy)-y*sin(2xy)∂Z/∂y=x*cos(xy)-2cos(

对x求导,e^z*z'(x)=yz+xyz'(x),z'(x)=yz/(e^z-xy)对y求导,e^z*z'(y)=xz+xyz'(y),z'(y)=xz/(e^z-xy)

再问：可以再帮我答题吗，我这边有很多财富值可以给你再问：

z=(x^2)*ln(2xy),Zx=(2x)ln(2xy)+(x^2)/2xy*(2xy)'=(2x)ln(2xy)+xZxx=2ln(2xy)+(2x)/2xy*(2xy)'+1=2ln(2xy)

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)

见下图

令a=x^2-y^2b=e^(xy)f具有一阶连续偏导数f1‘和f2’∂u/∂x=(∂u/∂a)×(∂a/∂x)+(∂

z'(x)=1/[1+(x^y)]*1/2√(x^y)*yx^(y-1)=yx^(y-1)/{2√(x^y)[1+(x^y)]}z'(y)=1/[1+(x^y)]*1/2√(x^y)*lnx*x^y=

z=f(xlny,x-y)əz/əx=lnyf1′+f2′əz/əy=(x/y)f1′-f2′再问：�жϼ��(n��1��)(-1)^n/��(n(

f1表示f对第1个变量求导数,其余类推.∂μ/∂x=f1+f2(y)+f3(yz+xy∂φ/∂x)=f1+yf2+y(z+x∂φ/ͦ

(1)z=ln(tanx/y)dz/dx=1/(tanx/y)*(sec²x/y)=sec²x/tanxdz/dy=1/(tanx/y)*(-tanx/y²)=-1/y(

(注:偏导数的符号姑且用"d"表示)dz/dx=1/{y[2(x/y)^0.5]}(算z对x的偏导数时,把y看成是一个常数即可)dz/dy=-x/{y^2*[2(x/y)^0.5]}(算z对y的偏导数