设z=uv sint,u=e,v=cost求
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由z=u²v²,其中u=x-y,v=x+y,题型:求复合函数的偏导数:z=(x-y)²(x+y)²,dz/dx=(x-y)²×2(x+y)+2(x-y
令u=x-y,v=y/xaz/ax=az/au×au/ax+az/av×av/ax=fu-y/x^2×fva^2z/axay=a(az/ax)/ay=a(fu-y/x^2×fv)/ay=a(fu)/a
Z=U*V则∂Z/∂U=V∂Z/∂V=UX=e^UsinV则∂X/∂U=e^UsinV=X∂X/∂V=e
∂z/∂x=(∂f(u,v)/∂u)*(∂u/∂x)+(∂f(u,v)/∂v)*(∂v/
z(x)+z(y)=-(f(x)+f(y))/f(z)f(x)=f1(1-z(x)-f2z(x))f(y)=-f1z(y)+f2(1-z(y))f(z)=-f1-f2所以z(x)+z(y)=1+z(x
x=ue^u两边微分:dx=e^udu+ue^udu=[(1+u)e^u]dudu/dx=1/[(1+u)e^u]u^2+v^2=1两边微分:2udu+2vdv=0dv/du=-u/vdv/dx=(d
z=(x+y)^2*cos(x^2*y^2)dz/dx=2*(x+y)*cos(x^2*y^2)-2*(x+y)^2*sin(x^2*y^2)*x*y^2dz/dy=2*(x+y)*cos(x^2*y
令e^xsiny=u,x^2+y^2=v则δz/δx=δf/δu*δu/δx+δf/δv*δv/δx=δf/δu*(e^xsiny)+δf/δv*(2x)δ^2z/δx^2=δ^2f/δu^2*(e^
∵z=f(x,xy),令u=x,v=xy∴∂z∂x=f′1+yf′2∴∂2z∂x∂y=∂∂y(f′1+yf′2)=∂f′1∂y+∂∂y(yf′2)═(∂f′1∂u∂u∂y+∂f′1∂v∂v∂y)+f′
其实就是求z的导数,cost^2求导为2cost*(-sint),t^6求导是6t^5,cost*t^3求导是-sint*t^3+cost*3*t^2,综合起来就是2cost*(-sint)+6t^5
dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(
2(x+y),2(x-y).下次弄个难点的
dz/dx=dz/du*(du/dx)=2u*1=2udz/dy=dz/du*(du/dy)=2u*1=2u和v没关系
说明:eu应该是e的x次幂,dz/dx,dz/dy应该是偏导数.∵v=xy,u=x2-y2∴du/dx=2x,du/dy=-2y,dv/dx=y,dv/dy=x∵z=ln(e^u+v),∴dz/dx=
dz/dx是z对x的偏导,这样把u,v都带入的话直接球偏导就好了dz/dx=y*e^(xy)*sin(x+y)+e^(xy)*cos(x+y)同理也可得到dz/dy=x*e^(xy)*sin(x+y)
∂z/∂x=∂z/∂u*du/dx+∂z/∂v*dv/dx=1/(u^2+v)*2u+1/(u^2+v)*2xy∂z
由柯西-黎曼条件v'(x)=-u'(y),v'(y)=u'(x)得u'(y)=-6xy,u'(x)=3y²-3x²因而选择B
z=u²v+3uv^4,u=e^x,v=sinx,求dz/dxdz/dx=2uu'v+u^2v'+3u'v^4+3v(4v^3)v'=2e^(2x)sinx+e^(2x)cosx+3e^x(