求函数[img=107x38]17da6537b12a2e0.png[/img]的导数； ( ) A: 2*x*sin(1/x) - sin(1/x) B: 2xsin(1/x) - cos(1/x) C: 2*x*sin(1/x) - cos(1/x) D: 2*x*cos(1/x) - cos(1/x)

函数 $y=sinx -x$的微分为 A: $cos x -1 $ B: $(cos x -1)dx$ C: $(sin x -1)dx$ D: $sin x -1$

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: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)

$\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ A: ${1 \over 4}\ln \left| {{{2\cos x + \sin x} \over {2\cos x - \sin x}}} \right| + C$ B: ${1 \over 4}\ln \left| {{{2\cos {x \over 2} + \sin {x \over 2}} \over {2\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ C: $\ln \left| {{{\cos {x \over 2} + \sin {x \over 2}} \over {\cos {x \over 2} - \sin {x \over 2}}}} \right| + C$ D: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$

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)

【单选题】在 MATLAB 命令窗口中 , 键入命令 syms x ; int(x*sin(x)) 。结果是 A. ans= sin(x)-x*cos(x) B. ans= cos(x)+x*sin(x) C. ans= sin(x)-cos(x) D. ans= -1/2*cos(x)*sin(x)+1/2*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)

cos(x)*cos(x/2)*cos(x/4)*cos(x/8).cos(x/(2^(n-1))

‍求以下定积分可以使用的命令有（）。‎‍[img=187x84]1802f8c916e3b88.jpg[/img]‎ A: I=integral(@(x) x./(1+abs(cos(x))),1/exp(1),exp(1)) B: f=@(x) x./(1+abs(cos(x))); q=integral(f,1/exp(1),exp(1)) C: f=@(x) x./(1+abs(cos(x))); q=integral(@f,1/exp(1),exp(1)) D: syms xf=x/(1+abs(cos(x)));s=int(f,x,1/exp(1),exp(1))eval(s)

函数\(y = \sin {1 \over x}\)的导数为（ ）. A: \({1 \over { { x^2}}}\sin {1 \over x}\) B: \( - {1 \over { { x^2}}}\sin {1 \over x}\) C: \( - {1 \over { { x^2}}}\cos {1 \over x}\) D: \({1 \over { { x^2}}}\cos {1 \over x}\)