设z=v/u,而u=x-2y,v=2x+y,求:偏z/偏x,偏z/偏y?

设z=u^v,而u=x+2y,v=2x+y,求全微分

az/ax=v·u^(v-1)·1+u^v·lnu·2
az/ay=v·u^(v-1)·2+u^v·lnu·1
所以
dz=[v·u^(v-1)·1+u^v·lnu·2]dx+[v·u^(v-1)·2+u^v·lnu·1]dy

∂u/∂x=cosy，∂u/∂y=-xsiny；
∂v/∂x=siny，∂v/∂y=xcosy；
∂z/∂u=2v.u^(2v-1);
∂z/∂v=2(lnu)u^2v
∂z/∂x=∂z/∂u.∂u/∂x+∂z/∂v.∂v/∂x
=2v.u^(2v-1).cosy+2(lnu)u^2v.siny
=2[(v/u)u^2v.cosy+(lnu)u^2v.siny]
=2z[(v/u)cosy+(lnu)siny]
=2z[siny+ln(xcosy)siny]
=2zsiny[1+ln(xcosy)]
=2zsinyln(excosy)
∂z/∂y=∂z/∂u.∂u/∂y+∂z/∂v.∂v/∂y
=2v.u^(2v-1).(-xsiny)+2(lnu)u^2v.xcosy
=2zx[-v/usiny+(lnu)cosy]
=2zx[-tanysiny+ln(xcosy)cosy]

z= v/u
= (2x+y)/(x-2y)
偏z/偏x = [(x-2y)2-(2x+y)]/(x-2y)^2
= -5y/(x-2y)^2
偏z/偏y= [(x-2y)-(2x+y)(-2)]/(x-2y)^2
=5x/(x-2y)^2

求下列函数的一阶偏导数,设z=uv,u=2x+y,v=x-2y.
答：=2(x-2y)+(2x+y)=4x-3y ∴αz/αy=αz/αu*αu/αy+αz/αv*αv/αy =v*1+u*(-2)=v-2u =(x-2y)-2(2x+y)=-3x-4y

求解:设z=uv,而u=e^t,v=cost,求dz/dt 谢谢呀!
答：∵z=uv，u=e^t，v=cost ∴z=(e^t)cost ∴dz/dt =[(e^t)cost]/dt =[(e^t)/dt]cost+[(cost)/dt](e^t)=(e^t)cost-(e^t)sint =(e^t)(cost-sint)=√2(e^t)(√2/2cost-√2/2sint)=√2(e^t)(cos45°cost-sin45°sint)=√2(e^t)cos(t+45°)先到(e^t)(c...

设z=ln(eu+v),v=xy,u=x2-y2,求dz/dx,dz/dy.?
答：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=[(e^u)/(e^u+v)](du/dx)+[1/(e^u+v)](dv/dx)=(2xe^u+y)/(e^u+v)dz/dy=[e^u/(e^u+v)](du/dy)+[1/(e^u+v...

设z=e^usinv 而u=xy v=x+y 求偏导数
答：1、对X的偏导数：Z'x=Z'u · u'x +Z'v · v'x =e^u sinv · y +e^u cosv · 1 =(ysinv+cosv) e^u 2、对Y的偏导数：Z'y=Z'u · u'y +Z'v · v'y =e^u sinv · x +e^u cosv · 1 =(xsinv+cosv) e^u 解法二：1、对X的偏导数：u^z=e^xy *(x...

设z=f(u+v)+Ψ(v),其中函数f,Ψ可微且x=u2+v2,y=u3-v3,求Z对X的偏导
答：1、本题的解答方法是：A、根据联立方程，求出u、v对x的偏导；B、然后对f(u+v)+Ψ(v)，运用链式求导法则；2、具体解答如下，若点击放大，图片更加清晰。

设Z=ulnv,u=x+y,v=x-y,求偏导数ZxZy
答：设z=u*ulnv,u=y/x,v=x*x y*y 则有 Z=y/x*y/xlnx*xy*y z=e^x+f(u,v)∂z/∂x=e^x+∂f/∂u·∂u/∂x+∂f/∂v·∂v/∂x ∂z/∂y=∂f/∂u·∂u/∂y+∂f...

设z=u^v,u=e^x,v=根号下(x^2+1),求dz/dx
答：dz/dx=(偏z/偏u)·du/dx+(偏z/偏v)·dv/dx =v·u^(v-1)·e^x+u^v·lnu·[1/2√(x²+1)]·2x =v·u^(v-1)·e^x+u^v·lnu·x/√(x²+1)再将u=e^x,v=√(x^2+1)代入，得：dz/dx=v·u^v+u^v·x·x/v =u^v·(v+x²/v)=[(e^x)^...

设z=u^2+v^2,u= x+ y,v=x-y,求dz
答：z'x=z'u*u'x+z'v*v'x =2u+2v z'y=z'u*u'y+z'v*v'y =2u-2v dz=z'xdx+z'ydy =(2u+2v)dx+(2u-2v)dy

设z=u^2+v^2,u=x^2-y,v=x+y^3,求αz/αx, αz/αy
答：简单计算一下即可，答案如图所示

设z=u^v,而u=x+2y,v=2x+y,求全微分
答：az/ax=v·u^(v-1)·1+u^v·lnu·2 az/ay=v·u^(v-1)·2+u^v·lnu·1 所以 dz=[v·u^(v-1)·1+u^v·lnu·2]dx+[v·u^(v-1)·2+u^v·lnu·1]dy