设\(z = {e^u}\sin v,\;u = xy,\;v = x + y\)，则\( { { \partial z} \over {\partial y}}=\)( )
A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\)
B: \(x{e^{xy}}\sin \left( {x + y} \right) \)
C: \( {e^{xy}}\cos \left( {x + y} \right)\)
D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)
A: \(x{e^{xy}}\sin \left( {x + y} \right) + {e^{xy}}\cos \left( {x + y} \right)\)
B: \(x{e^{xy}}\sin \left( {x + y} \right) \)
C: \( {e^{xy}}\cos \left( {x + y} \right)\)
D: \(x{e^{xy}}\sin \left( {x + y} \right) - {e^{xy}}\cos \left( {x + y} \right)\)
举一反三
- 设\(z = xy{e^{\sin xy}}\)，则\({z'_y} = \)（ ）。 A: \(x{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) B: \(y{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) C: \(x{e^{\sin xy}}\left( {1 + y\cos xy} \right)\) D: \(x{e^{\sin xy}}\left( {1 - xy\cos xy} \right)\)
- 设\(z = z\left( {x,y} \right)\)是由方程\({z^3}{\rm{ + }}3xyz - 3\sin xy = 1\)确定的隐函数，则\( { { \partial z} \over {\partial y}}=\)( ) A: \( { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) B: \( { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\) C: \( { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) D: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)
- 函数\(z = {\left( {xy} \right)^x}\)的全微分为 A: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + x{\left( {xy} \right)^x}dy\) B: \(dz = \left( { { {\left( {xy} \right)}^x} + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) C: \(dz = {\left( {xy} \right)^x}\ln xydx + { { x { { \left( {xy} \right)}^x}} \over y}dy\) D: \(dz = {\left( {xy} \right)^x}\left( {1 + \ln xy} \right)dx + { { x { { \left( {xy} \right)}^x}} \over y}dy\)
- 下列方程中是线性微分方程的是（ ）。 A: \( \cos \left( {y'} \right) + {e^y} = x \) B: \( xy'' + 2y' - {x^2}y = {e^x} \) C: \( {\left( {y'} \right)^2} + 5y = 0 \) D: \( y'' + \sin y = 8x \)
- 已知\( y = {x^{\cos x}} \) ,则\( y' = \left( { - \sin x\ln x + { { \cos x} \over x}} \right){x^{\cos x}} \)( ).