z=e^x² y² 的偏导数-搜答案-智慧作业帮

/>d(sinxy-2z+e^z)=0dsinxy-d2z+de^z=0ycosxydx+xcosxydy-2dz+e^zdz=0ycosxydx+xcosxydy=2dz-e^zdz=(2-e^z)

Z=U*V则∂Z/∂U=V∂Z/∂V=UX=e^UsinV则∂X/∂U=e^UsinV=X∂X/∂V=e

G[x+z*y^(-1),y+z*x^(-1)]=0证明x*∂z/∂x+y*∂z/∂y=z-xy?Gz=(1/y)G1+(1/x)G2=LGx=G1-(

∂z/∂x只对x求导数,而把y看作一个常数,∂z/∂x=(x+y)'sin(x-y)+(x+y)sin(x-y)'=sin(x-y)+(x+y)cos(

z=xy+x/y对x的偏导数=y+1/y对y的偏导数=x-x/y^2

两边对z微分e^zdz-d（xyz）=0=e^zdz-xydz-zd（xy）=e^zdz-xydz-zxdy-zydx所以,整理两边：（e^z-xy）dz=zxdy+zydx所以：dz=zx/（e^z

1、对X求导(导数符号无,用“￡”代替)两边对x求导有：2x2z￡z/￡x=-ycos(z/x)/x^2*￡z/￡x:化简得：￡z/￡x=-2x/[2zycos(z/x)/x^2]:2、对y求导两边求

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(

求偏导时就是把其他变量当做常数.所以,对x的偏导为y*x^(y-1),对y的偏导是x^y*lnx.

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)

不需要图,很简单的z=xy+u两边对x求导：∂z/∂x=y+∂u/∂x,两边对y求导：∂²z/(∂x∂y)

把y看做常数∂z/∂x=e^(x²y)*(x²y)'=2yxe^(x²y)

z=e^xsiny-3(x^3)cosyzx=e^xsiny-9(x^2)cosyzy=e^xcosy+3(x^3)siny

1.z'x=3x²y²z'y=2x³y2.z'x=4x³z'y=3y³3.z'x=ye^(xy)+2xyz'y=xe^(xy)+x²4.u'

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)

xyz=e^(x+y)两边求关于x的偏导数(把z当成常数)∂(xyz)/∂x=∂e^(x+y)/∂xz∂(xy)/∂x=e^(x

your answer here

z=y/f(x²+y²),令u=x²+y²∂z/∂x=y·-1·[∂f(u)/∂u·∂(x²

dz/dx=f'（x+y,x-y）dz/dy=f'（x+y,x-y）(-1)=-f'（x+y,x-y）