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: \({e^{xy}}({x^2}y + {y^2} + 2x)\)
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: \({e^{xy}}({x^2}y + {y^2} + 2x)\)
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\(z = u{e^v}\)，\(u = x + y\)，\(v = xy\)，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: \({e^{xy}}(y+ xy + {y^2})\)
- 分解因式()x()3()y()-()2()x()2()y()2()+()xy()3()正确的是A.()xy()(()x()+()y())()2()B.()xy()(()x()2()﹣()2()xy()+()y()2())()C.()xy()(()x()2()+2()xy()﹣()y()2())()D.()xy()(()x()﹣()y())()2
- 9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定，则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$（ ） A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
- 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: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)