求函数[img=192x40]17da653862ff7b6.png[/img]的导数; ( ) A: cos(x)/sin(x) - cot(x)*(cot(x)^2 + 1) B: cos(x)/sin(x) C: cot(x)*(cot(x)^2 + 1) D: cos(x)/sin(x) - cot(x)*(cot(x)^2 + 1)+cot(x)
求函数[img=192x40]17da653862ff7b6.png[/img]的导数; ( ) A: cos(x)/sin(x) - cot(x)*(cot(x)^2 + 1) B: cos(x)/sin(x) C: cot(x)*(cot(x)^2 + 1) D: cos(x)/sin(x) - cot(x)*(cot(x)^2 + 1)+cot(x)
$\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ A: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\tan x + C$ B: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\cot x + C$ C: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\cot x + C$ D: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\tan x + C$
$\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ A: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\tan x + C$ B: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\cot x + C$ C: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\cot x + C$ D: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\tan x + C$
求极限LIM(趋近与0)(cot^2*x-1/x^2)
求极限LIM(趋近与0)(cot^2*x-1/x^2)
下列语句语法正确的是( ) A: if x<2*y and x>y then y=x**2 B: if x<2*y : x>y then y=x^2 C: if x<2*y and x>y then y=x2 D: if x<2*y and x>y then y=x^2
下列语句语法正确的是( ) A: if x<2*y and x>y then y=x**2 B: if x<2*y : x>y then y=x^2 C: if x<2*y and x>y then y=x2 D: if x<2*y and x>y then y=x^2
【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )
【单选题】对任意实数x 1 , y 1 , x 2 , y 2 , x 1 < x 2 , y 1 < y 2 , 分布函数P{x 1 <X≤x 2 , y 1 <Y≤y 2 }=? A. F(x 2 , y 2 )+ F(x 1 , y 1 )+ F(x 1 , y 2 )+ F(x 2 , y 1 ) B. F(x 2 , y 2 )- F(x 1 , y 1 )+ F(x 1 , y 2 )- F(x 2 , y 1 ) C. F(x 2 , y 2 )+ F(x 1 , y 1 )- F(x 1 , y 2 )- F(x 2 , y 1 ) D. F(x 2 , y 2 )- F(x 1 , y 1 )- F(x 1 , y 2 )+ F(x 2 , y 1 )
4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$,并且$z(x,0)=x,z(0,y)={{y}^{2}}$,则$z(x,y)=$( ) A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
已知\( y = \ln (\sin x) \),则\( y' \)为( ). A: \( {1 \over {\sin x}} \) B: \( {1 \over {\cos x}} \) C: \( \cot x \) D: \( - \cot x \)
已知\( y = \ln (\sin x) \),则\( y' \)为( ). A: \( {1 \over {\sin x}} \) B: \( {1 \over {\cos x}} \) C: \( \cot x \) D: \( - \cot x \)
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
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
求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]