• 2022-06-08
    设方程\({x^2} + {y^2} + {z^2} = 2Rx\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial x}}=\)
    A: \( { { \partial z} \over {\partial x}} = { { R +x} \over z}\)
    B: \( { { \partial z} \over {\partial x}} =- { { R +x} \over z}\)
    C: \( { { \partial z} \over {\partial x}} = { { R - x} \over z}\)
    D: \( { { \partial z} \over {\partial x}} =- { { R - x} \over z}\)
  • C

    内容

    • 0

      由方程\({z^3} - 3xyz = {a^3}\)所确定的隐函数\(z= f(x,y)\)的偏导数\( { { \partial z} \over {\partial x}} = \) A: \( { { yz} \over { { z^2} - xy}}\) B: \(- { { yz} \over { { z^2} - xy}}\) C: \( { { yz} \over { { z^2} +xy}}\) D: \(- { { yz} \over { { z^2}+xy}}\)

    • 1

      设方程\(\sin z - xyz = 0\)确定函数\(z=z(x,y)\),则\( { { \partial z} \over {\partial y}}=\)( )。 A: \( { { xz} \over {xy+cos z }}\) B: \(- { { xz} \over {xy+cos z }}\) C: \(- { { xz} \over {\cos z - xy}}\) D: \( { { xz} \over {\cos z - xy}}\)

    • 2

      设\(z = z\left( {x,y} \right)\)是由方程\({z^3}{\rm{ + }}3xyz - 3\sin xy = 1\)确定的隐函数,则\( { { \partial z} \over {\partial y}}=\)( ) A: \( { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) B: \( { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\) C: \( { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) D: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)

    • 3

      设\(z = {u^2}{\rm{ + }}{v^2}\),\(u = x + y\),\(v = x - y\),则\( { { \partial z} \over {\partial x}}=\) A: \(4y\) B: \(4x\) C: \(2(x+y)\) D: \(2(x-y)\)

    • 4

      \[设z=z(x,y)是由方程f(cx-az,cy-bz)=0所确定的函数,则a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=()\] A: a B: b C: c D: 0