设方程\({sinz} - x^2yz = 0\)确定函数\(z=z(x,y)\)，则\( { { \partial z} \over {\partial x}}=\) A: \( { { 2xyz} \over {\cos z - {x^2}y}}\) B: \( { { 2xyz} \over {\cos z + {x^2}y}}\) C: \( { { xyz} \over {\cos z - {x^2}y}}\) D: \( { { 2xy} \over {\cos z - {x^2}y}}\)

设\(z = u{e^v}\)，\(u = {x^2} + {y^2}\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}({x^2}y + {y^3} + 2x)\) B: \({e^{xy}}({x}y^2 + {y^3} + 2x)\) C: \({e^{xy}}({x}y + {y^3} + 2x)\) D: \({e^{xy}}({x^2}y + {y^2} + 2x)\)

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: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$

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: $1+x+\frac{1}{2}({{x}^{2}}+{{y}^{2}})$

已知函数由下列方程确定$x^2 - y^2=1 $，则$\frac{d^2 y}{d^2 x} =$( )。 A: $\frac{1}{y^2}$ B: $-\frac{1}{y^2}$ C: $-\frac{1}{y^3}$ D: $\frac{1}{y^3}$