设\(z = u{e^v}\)，\(u = x + y\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\)
A: \({e^{xy}}(1 + xy + {y^2})\)
B: \({e^{xy}}(1 + xy + {y^3})\)
C: \({e^{xy}}(x+ xy + {y^2})\)
D: \({e^{xy}}(y+ xy + {y^2})\)
A: \({e^{xy}}(1 + xy + {y^2})\)
B: \({e^{xy}}(1 + xy + {y^3})\)
C: \({e^{xy}}(x+ xy + {y^2})\)
D: \({e^{xy}}(y+ xy + {y^2})\)
举一反三
- 设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial y}}=\)（ ）。 A: \({e^{xy}}({x}y^2 + {x^3} + 2y)\) B: \({e^{xy}}({x^2}y + {x^3} + 2y)\) C: \({e^{xy}}({x}y^2 + {x^3} + 2x)\) D: \({e^{xy}}({x}y+ {x^3} + 2y)\)
- 设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}({x^2}y + {y^3} + 2x)\) B: \({e^{xy}}({x}y^2 + {y^3} + 2x)\) C: \({e^{xy}}({x}y + {y^3} + 2x)\) D: \({e^{xy}}({x^2}y + {y^2} + 2x)\)
- 设\(z = {e^u}\sin v,\;u = xy,\;v = x + y\)，则\( { { \partial z} \over {\partial y}}=\)( ) A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\) B: \(x{e^{xy}}\sin \left( {x + y} \right) \) C: \( {e^{xy}}\cos \left( {x + y} \right)\) D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)
- 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: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)
- 设\(z = xy{e^{\sin xy}}\)，则\({z'_y} = \)（ ）。 A: \(x{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) B: \(y{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) C: \(x{e^{\sin xy}}\left( {1 + y\cos xy} \right)\) D: \(x{e^{\sin xy}}\left( {1 - xy\cos xy} \right)\)