• 2022-06-17
    函数y=2x2x2+1的导数是(  )
    A: y′=4x(x2+1)-4x2(x2+1)2
    B: y′=4x(x2+1)-4x3(x2+1)2
    C: y′=4x(x2+1)+4x3(x2+1)2
    D: y′=4x(x2+1)-4x(x2+1)2
  • B

    内容

    • 0

      将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)

    • 1

      已知\( y = {x^2} + 4x \),则\( dy \)为( ). A: \( (2x + 4)dx \) B: \( 2xdx \) C: \( ({x^2} + 4)dx \) D: \( ({x^2} + 4x)dx \)

    • 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}]

    • 3

      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}]

    • 4

      设X是取自总体 N(μ,σ2)的一个样本,μ 的无偏估计量是( ) A: 1/4·X1+2/4·X2 B: 2/4·X1+1/4·X2+1/4·X3 C: 1/4·X1+1/4·X2+1/4·X3 D: 2/4·X1+2/4·X2+1/4·X3