举一反三
- 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\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，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: \({e^{xy}}({x}y+ {x^3} + 2y)\)
- 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
- 已知\( y = {x^2} + 4x \)，则\( dy \)为（ ）. A: \( (2x + 4)dx \) B: \( 2xdx \) C: \( ({x^2} + 4)dx \) D: \( ({x^2} + 4x)dx \)
- 已知\( {y^{(4)}} = {x^2} + 2x \)，则\( {y^{(5)}} = 2x + 2 \)（ ）.
内容
- 0
已知\( y = {x^3}\cos 2x \)，则\( y'' \)为（ ）. A: 0 B: \( 6x\cos 2x{\rm{ + }}12{x^2}\sin 2x - 4{x^3}\cos 2x \) C: \( 6x\cos 2x - 12{x^2}\sin 2x{\rm{ + }}4{x^3}\cos 2x \) D: \( 6x\cos 2x - 12{x^2}\sin 2x - 4{x^3}\cos 2x \)
- 1
设全体域D是正整数集合，确定下列命题的真值： (1) "x$y (xy=y) ( ) (2) $x"y(x+y=y) ( ) (3) $x"y(x+y=x) ( ) (4) "x$y(y=2x) ( )
- 2
函数\(y = \sin{x^2}\)的导数为( ). A: \( - 2x\sec {x^4}\) B: \(2x\cos {x^2}\) C: \(2x\sec {x^2}\) D: \(- 2x\sec {x^2}\)
- 3
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: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$
- 4
在空间直角坐标系中，下面表示平面方程的是（ ）. A: \( {x^2} + {y^2} + {z^2} = 4 \) B: \( 2x - 6y + 2z - 7 = 0 \) C: \( 3{x^2} + 4{y^2} = 1 \) D: \( 4{y^2} + \frac { { {z^2}}}{3} = 1 \)