【单选题】化简 sin( x + y )sin( x - y ) + cos( x + y )cos( x - y ) 的结果是 A. sin 2 x B. cos 2 y C. － cos 2 x D. -cos 2 y

曲线积分$$\int_{(0,0}^{(x,y)}(2x\cos y-y^2\sin x)dx+(2y\cos x-x^2\sin y)dy=$$ A: $y^2\cos x+x^2\cos y$ B: $x^2\cos x+y^2\cos y$ C: $x^2\sin y+y^2\sin x$ D: $x^2\sin x+y^2\sin y$

3. $(2x\cos y-{{y}^{2}}\sin x)dx+(2y\cos x-{{x}^{2}}\sin y)dy$的原函数是 ( ) A: ${{x}^{2}}\sin y-{{y}^{2}}\sin x+C$ B: ${{x}^{2}}\sin y+{{y}^{2}}\sin x+C$ C: ${{x}^{2}}\cos y-{{y}^{2}}\cos x+C$ D: ${{x}^{2}}\cos y+{{y}^{2}}\cos x+C$

在区域<img src="http://img2.ph.126.net/gYKpVz-ihv2c735JpXLFXA==/148900262780427590.png" />画出函数<img src="http://img1.ph.126.net/1MufBuQ2l1d-e3P0WpTtOA==/6597356739194210350.png" />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}]

以下表达式中，有两个的计算结果是相同的，请挑选出来 A: 1 / sqrt(sin(x) * sin(x) + cos(y) * cos(y)) B: sqrt(pow(sin(x), 2) + pow(cos(y), 2)) C: pow(sin(x) * sin(x) + cos(y) * cos(y), 0.5) D: pow(pow(sin(x), 2) + pow(cos(y), 2), 2)

设方程\({sinz} - x^2yz = 0\)确定函数\(z=z(x,y)\)，则\( { { \partial z} \over {\partial x}}=\) A: \( { { 2xyz} \over {\cos z - {x^2}y}}\) B: \( { { 2xyz} \over {\cos z + {x^2}y}}\) C: \( { { xyz} \over {\cos z - {x^2}y}}\) D: \( { { 2xy} \over {\cos z - {x^2}y}}\)

求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)

【单选题】设y=sin(cos(x)),求 结果为:(本题10.0分) A. cos(cos(x))*cos(x)+ sin(cos(x))*sin(x)^2 B. - cos(cos(x))*cos(x) - sin(cos(x))*sin(x)^2 C. - cos(cos(x))*cos(x)^2 - sin(cos(x))*sin(x)^2 D. - cos(cos(x))*cos(x) ^2- sin(cos(x))*sin(x)

已知 \( y = \sin x + \ln 2 \)，则 \( y' = \cos x + {1 \over 2} \)（ ）.

已知\( {y^{(n)}} = \cos x \)，则\( {y^{(n + 2)}} \)为（ ）. A: \( \sin x \) B: \( - \sin x \) C: \( \cos x \) D: \( - \cos x \)