y=sin（2x+1）,dy=?
A: 2cos（2x+1）dx
B: cos（2x+1）dx
A: 2cos（2x+1）dx
B: cos（2x+1）dx
举一反三
- y=cos(2x+1)的微分是dy=-sin(2x+1)dx
- 下列方程中( )是一阶线性微分方程。 A: \( 2{x^2}yy' = {y^2} + 1 \) B: \( xy' + {y \over x} - x = 0 \) C: \( \cos y + x\sin y { { dy} \over {dx}} = 0 \) D: \( y'' + xy' = 4{x^2} + 1 \)
- $\int {{1 \over {3 + 5\cos x}}} dx = \left( {} \right)$ 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: $\ln \left| {{{\cos x + \sin x} \over {\cos x - \sin x}}} \right| + C$
- $\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ 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$
- 函数 $y=sinx -x$的微分为 A: $cos x -1 $ B: $(cos x -1)dx$ C: $(sin x -1)dx$ D: $sin x -1$