曲线y=e2x在x=2处切线的斜率是()
A: 2e4
B: e2
C: 2e2
D: 2
A: 2e4
B: e2
C: 2e2
D: 2
举一反三
- 设二维随机变量 (X , Y )服从二维正态分布，则随机变量X + Y与X – Y不相关的充要条件为（ ） A: E (X ) = E (Y ) B: E (X 2) – [E (X )]2 = E (Y 2 ) – [E (Y )]2 C: E (X 2 ) = E (Y 2) D: E (X 2) + [E (X )]2 = E (Y 2 ) + [E (Y )]2
- 过曲线\(y = {x^2}\)上点\((2,4)\)处切线的斜率为( ). A: 2 B: 4 C: 6 D: 8
- 方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 方程$(x^2+1)(y^2-1) + xy y' = 0$的通解为 A: $y^2 = C \frac{e^{-x^2}}{x^2}$ B: $y = C \frac{e^{-x^2}}{x^2}$ C: $y^2 = C \frac{e^{-x^2}}{x^2}+1$ D: $y=C \frac{e^{-x^2}}{x^2}+1$
- 已知\( y = {x^2}{e^{ - x}} \)，则\( y'' \)为（ ）. A: \( 2{e^{ - x}} - 4x{e^{ - x}} - {x^2}{e^{ - x}} \) B: \( 2{e^{ - x}} - 4x{e^{ - x}} + {x^2}{e^{ - x}} \) C: 0 D: \( 2{e^{ - x}} - 4x{e^{ - x}} \)