在区域画出函数的密度图形。? ContourPlot[Sin[x^2]+y Cos[y^2],{x,-6,6},{y,-6,6}]|DensityPlot[Sin[x^2]+y Cos[y^2],{x,-6,6},{y,-6,6}]|DensityPlot[Sin[x^2]+y Cos[y^2],{x,-6,6},{y,-6,6}]|ContourPlot[Sin[x^2]+y Cos[y^2],{x,-6,6},{y,-6,6}]
举一反三
- 画抛物柱面[img=51x26]180307326eca386.png[/img],画图区域: [img=260x25]1803073277226f9.png[/img] A: ContourPlot3D[x=y^2,{x,-6,6},{y,-3,3},{z,-5,5}] B: ContourPlot3D[x=y^2,{y,-3,3},{z,-5,5},{x,-6,6}] C: ContourPlot3D[x==y^2,{x,-6,6},{y,-3,3},{z,-5,5}] D: ContourPlot3D[x==y^2,{y,-3,3},{z,-5,5},{x,-6,6}]
- 应用Matlab软件计算行列式[img=110x88]17da5d7b00219d6.png[/img]为( ). A: x^2 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 B: x^3 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 C: x^4 - 6*x^2*y^2 + 8*x*y^3 - 3*y^4 D: x^5- 6*x^2*y^2 + 8*x*y^3 - 3*y^4
- Let $y=x^{2}\sin{x}$ Solve $y^{(6)}(0)=$:<br/>______
- 已知\( y = \ln (6 - {x^2}) \),则\( y' \)为( ). A: \( { { 2x} \over {6 - {x^2}}} \) B: \( { { - 2x} \over {6 - {x^2}}} \) C: \( {1 \over {6 - {x^2}}} \) D: \( { { {x^2}} \over {6 - {x^2}}} \)
- 已知\( {y^{(6)}} = \cos x \),则\( {y^{(8)}} = - \sin x \)( ).