函数$f(x,y)={{\text{e}}^{-x}}\cos y$在点$(0,0)$处2次Taylor多项式为
A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$
B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$
C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
A: $1+x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$
B: $1-x+\frac{1}{2}({{x}^{2}}-{{y}^{2}})$
C: $1-x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
D: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$
举一反三
- 函数$f(x,y)={{\text{e}}^{-x}}\ln (1+y)$在点$(0,0)$处2次Taylor多项式为 A: $y+\frac{1}{2}(-2xy-{{y}^{2}})$ B: $y+\frac{1}{2}(-2xy+{{y}^{2}})$ C: $y+\frac{1}{2}(2xy-{{y}^{2}})$ D: $y+\frac{1}{2}(-xy-{{y}^{2}})$
- 4.已知二元函数$z(x,y)$满足方程$\frac{{{\partial }^{2}}z}{\partial x\partial y}=x+y$，并且$z(x,0)=x,z(0,y)={{y}^{2}}$，则$z(x,y)=$（ ） A: $\frac{1}{2}({{x}^{2}}y-x{{y}^{2}})+{{y}^{2}}+x$ B: $\frac{1}{2}({{x}^{2}}{{y}^{2}}+xy)+{{y}^{2}}+x$ C: ${{x}^{2}}{{y}^{2}}+{{y}^{2}}+x$ D: $\frac{1}{2}({{x}^{2}}y+x{{y}^{2}})+{{y}^{2}}+x$
- 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: $y=C \frac{e^{-x^2}}{x^2}+1$
- 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: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 下列函数在点$(0,0)$的重极限存在的是 A: $f(x,y)=\frac{y^2}{x^2+y^2}$ B: $f(x,y)=(x+y)\sin\frac{1}{x}\sin\frac{1}{y}$ C: $f(x,y)=\frac{x^2y^2}{x^2y^2+(x-y)^2}$ D: $f(x,y)=\frac{x^2y^2}{x^3+y^3}$