设\(z = \int_ { { x^2}}^y { { e^t}\sin t} dt\)，则\({z_{xx}=}\) A: \(2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) B: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} - 2{x^2}\cos {x^2}} \right]\) C: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\sin {x^2} + 2{x^2}\cos {x^2}} \right]\) D: \( - 2{e^ { { x^2}}}\left[ {\left( {1 + 2{x^2}} \right)\cos {x^2} + 2{x^2}\sin {x^2}} \right]\)

求不定积分[img=132x48]17da6537fc8dad6.png[/img]； （ ） A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))

与数学关系式[img=114x22]1803bce8722f322.png[/img]等价的C语言关系表达式是？ A: x < -2 && x > 2 B: x < -2 || x > 2 C: -2 < x < 2 D: !(-2 <= x <=2) E: !(-2 <=x && x <= 2) F: x < -2, x > 2

求函数[img=148x49]17da6537a5eee98.png[/img]的导数； ( ) A: 1/(x^2*(2/x^2 + 1)) B: -1/(x^2*(2/x^2 + 1)) C: (x^2*(2/x^2 + 1)) D: -1/(x^2*(2/x^2 + 1))+2/x^2 + 1

求函数[img=102x46]17da6537bc771a0.png[/img]的导数； ( ) A: -x/(a^2 - x^2)^(3/2) B: x/(a^2 - x^2)^(3/2) C: (a^2 - x^2)^(3/2)/x D: (a^2 - x^2)^(1/2)

求微分方程[img=101x35]17da5f15503f795.png[/img] 的通解,实验命令为(). A: dsolve(Dy+2*x*y=x*exp(-x^2))ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 B: dsolve('Dy+2*x*y=x*exp(-x^2)','x')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2 C: dsolve('Dy+2*x*y=x*exp(-x^2)')ans=C1*exp(-x^2) + (x^2*exp(-x^2))/2

与数学关系式[img=114x22]17de85f03ec5890.png[/img]等价的C语言关系表达式是？ A: x <; -2 &amp;&amp; x >; 2 B: x <; -2 || x >; 2 C: -2 <; x <; 2 D: !(-2 <;= x <;=2) E: !(-2 <;=x &amp;&amp; x <;= 2) F: x <; -2, x >; 2

数学式 A: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sinx^2-Cos2x)) B: (Exp(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) C: (Exp(2*x)*Ln(x)+x^2)/Sqr(Abs(Sin(x^2)-Cos(x)^2)) D: (e^(2*x)*Log(x)+x^2)/Sqr(Abs(Sin(x)^2-Cos(x)^2))

已知\( y = f({x^2}) \)，假设\( f(u) \)二阶可导，则\( y'' \)为（ ）. A: \( 4{x^2}f''({x^2}){\rm{ + }}2f'({x^2}) \) B: \( {x^2}f''({x^2}){\rm{ + }}2f'({x^2}) \) C: \( 4{x^2}f''({x^2}){\rm{ + }}f'({x^2}) \) D: \( {x^2}f''({x^2}){\rm{ + }}f'({x^2}) \)

‏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}]