设z=u^v,u=2+sin(xy),v=x-2y,求δz/δx,δz/δy
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设二阶偏导数连续的函数z=f(u,v),u=xy,v=x^2-y^2,求δz/δx,δz/δy,最好有详细的步骤
即z=e^ [ln(2+sinxy) *(x-2y)]
那么求偏导数得到
δz/δx=z *δ[ln(2+sinxy) *(x-2y)]/δx
=z *[y* cosxy/(2+sinxy) *(x-2y) +ln(2+sinxy)]
δz/δy=z *δ[ln(2+sinxy) *(x-2y)]/δy
=z *[x *cosxy/(2+sinxy) *(x-2y) -2ln(2+sinxy)]
设f(u,v)为二元可微函数,Z=f(sin(x+y),exy),则?z?x=__
答:设u=sin(x+y),v=exy,则z=f(u,v)∴?z?x=?f?u?u?x+?f?v?v?x=cos(x+y)?f?u+yexy?f?v=cos(x+y)f1′+yexyf2′
设z=e^usinv 而u=xy v=x+y 求偏导数
答:=(ysinv+cosv) e^u Z'y=Z'u · u'y +Z'v · v'y =e^u sinv · x +e^u cosv · 1 =(xsinv+cosv) e^u 或 ^^^z=e^xy *(x+y)那么对x求偏导数得到 Z'x=(e^xy)' *(x+y)+e^xy *(x+y)'=y *e^xy *(x+y)+e^xy =e^xy *(xy+y^2+1)同理Z'y=...
高等数学
答:= 2[t+sin2t/2]<0,π/2> = π.三 1. z=xe^(xy), z'<y>=x^2e^(xy), z''<yy>=x^3e^(xy).2. z=f(x^2y,xy^2), 令 u=x^2y, v=xy^2,z'<x>=f'<u>u'<x>+f'<v>v'<x> = 2xyf'<u>+y^2f'<v>.3. 令 t=√x,则 ∫<0,+∞>e^(-√x)dx = ...
z=(e^u)sinv,u=xy^2,v=x-y,求dz
答:dz=(e^u)sinvdu+(e^u)cosvdv =(e^u)sinv[y²dx+2xydy]+(e^u)cosv[dx-dy]=e^(xy²)[y²sin(x-y)+cos(x-y)]dx+e^(xy²)[2xysin(x-y)-cos(x-y)]dy
u=sin(x+yz),其中x=uv,y=u+v,z=e的u次方
答:du=cos(x+yz)[vdu+udv+e^u*(du+dv)+(u+v)e^u*du],整理得{1-cos(x+yz)[v+e^u+(u+v)e^u]}du=cos(x+yz)(u+e^u)dv,所以du/dv=(u+e^u)cos(x+yz)/{1-[v+(1+u+v)e^u]cos(x+yz)}.
两道高等数学题,求详细解答
答:方法如下,请作参考:
设z=(e^u)*sinv,u=xy,v=x+y,求δz/δx和δz/δy.
答:δz/δx=(δz/δu)*(δu/δx)+(δz/δv)*(δv/δx)=((e^u)*sinv)*y+((e^u)*cosv)*1 =(e^xy)*[ysin(x+y)+cos(x+y);δz/δy=(δz/δu)*(δu/δy)=(δz/δv)*(δv/δy)=((e^u)*sinv)*x+((e^u)*cosv)*1 =(e^xy)*[xsin(x+y)+cos(x+...
设z=u²-v²,且u=cosx,v=sinx,求dz/dx?
答:详细完整过程rt所示……希望能帮到你解决问题
z=e的u次方sin(uv),其中 u= xy, v= x+ y,求 ∂z/ ∂x;∂z /...
答:貌似答案确实写的有些问题 这里是求偏导的链式法则 最后一步没有进行 ðu/ðx和ðu/ðy 那么各项相乘之后 就缺少了一个乘以y和乘以x 应该像你那样添上这一步才行
z=u²-v²,u=xcosy,v=ysinx,求∂z/∂x和∂z/∂y?
答:z的x偏导数是z的u偏导乘以u的x偏导+z的v偏导乘以v的x偏导。所以结果是z'(x)=2ucosy-2vycosx=2x(cosy)^2-2y^2sinxcosx=2x(cosy)^2-y^2sin2x z的y偏导数也是一个道理的 z'(y)=-2uxsiny-2vsinx=-2x^2sinycosy-2y(sinx)^2 =-x^2sin2y-2y(sinx)^2 ...
z=f(u,v), u=xy, v=x^2-y^2
du/dx=y, du/dy=x
dv/dx=2x, dv/dy=-2y
dz/dx=dz/du*du/dx+dz/dv*dv/dx
=df/du*y+df/dv*2x
dz/dy=dz/du*du/dy+dz/dv*dv/dy
=df/du*x-df/dv*2y
δz/δx=(δz/δu)*(δu/δx)+(δz/δv)*(δv/δx)
=((e^u)*sinv)*y+((e^u)*cosv)*1
=(e^xy)*[ysin(x+y)+cos(x+y);
δz/δy=(δz/δu)*(δu/δy)=(δz/δv)*(δv/δy)
=((e^u)*sinv)*x+((e^u)*cosv)*1
=(e^xy)*[xsin(x+y)+cos(x+y);
简单计算一下即可,答案如图所示
即z=e^ [ln(2+sinxy) *(x-2y)]
那么求偏导数得到
δz/δx=z *δ[ln(2+sinxy) *(x-2y)]/δx
=z *[y* cosxy/(2+sinxy) *(x-2y) +ln(2+sinxy)]
δz/δy=z *δ[ln(2+sinxy) *(x-2y)]/δy
=z *[x *cosxy/(2+sinxy) *(x-2y) -2ln(2+sinxy)]
答:设u=sin(x+y),v=exy,则z=f(u,v)∴?z?x=?f?u?u?x+?f?v?v?x=cos(x+y)?f?u+yexy?f?v=cos(x+y)f1′+yexyf2′
答:=(ysinv+cosv) e^u Z'y=Z'u · u'y +Z'v · v'y =e^u sinv · x +e^u cosv · 1 =(xsinv+cosv) e^u 或 ^^^z=e^xy *(x+y)那么对x求偏导数得到 Z'x=(e^xy)' *(x+y)+e^xy *(x+y)'=y *e^xy *(x+y)+e^xy =e^xy *(xy+y^2+1)同理Z'y=...
答:= 2[t+sin2t/2]<0,π/2> = π.三 1. z=xe^(xy), z'<y>=x^2e^(xy), z''<yy>=x^3e^(xy).2. z=f(x^2y,xy^2), 令 u=x^2y, v=xy^2,z'<x>=f'<u>u'<x>+f'<v>v'<x> = 2xyf'<u>+y^2f'<v>.3. 令 t=√x,则 ∫<0,+∞>e^(-√x)dx = ...
答:dz=(e^u)sinvdu+(e^u)cosvdv =(e^u)sinv[y²dx+2xydy]+(e^u)cosv[dx-dy]=e^(xy²)[y²sin(x-y)+cos(x-y)]dx+e^(xy²)[2xysin(x-y)-cos(x-y)]dy
答:du=cos(x+yz)[vdu+udv+e^u*(du+dv)+(u+v)e^u*du],整理得{1-cos(x+yz)[v+e^u+(u+v)e^u]}du=cos(x+yz)(u+e^u)dv,所以du/dv=(u+e^u)cos(x+yz)/{1-[v+(1+u+v)e^u]cos(x+yz)}.
答:方法如下,请作参考:
答:δz/δx=(δz/δu)*(δu/δx)+(δz/δv)*(δv/δx)=((e^u)*sinv)*y+((e^u)*cosv)*1 =(e^xy)*[ysin(x+y)+cos(x+y);δz/δy=(δz/δu)*(δu/δy)=(δz/δv)*(δv/δy)=((e^u)*sinv)*x+((e^u)*cosv)*1 =(e^xy)*[xsin(x+y)+cos(x+...
答:详细完整过程rt所示……希望能帮到你解决问题
答:貌似答案确实写的有些问题 这里是求偏导的链式法则 最后一步没有进行 ðu/ðx和ðu/ðy 那么各项相乘之后 就缺少了一个乘以y和乘以x 应该像你那样添上这一步才行
答:z的x偏导数是z的u偏导乘以u的x偏导+z的v偏导乘以v的x偏导。所以结果是z'(x)=2ucosy-2vycosx=2x(cosy)^2-2y^2sinxcosx=2x(cosy)^2-y^2sin2x z的y偏导数也是一个道理的 z'(y)=-2uxsiny-2vsinx=-2x^2sinycosy-2y(sinx)^2 =-x^2sin2y-2y(sinx)^2 ...