VC1.13

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Given,f=\cos xyi+(3xy-2x^{2})j-(3x+2y)k\,

Therefore,{\frac  {\partial f}{\partial x}}={\frac  {\partial }{\partial x}}(\cos xy)i+{\frac  {\partial }{\partial x}}(3xy-2x^{2})j-{\frac  {\partial }{\partial x}}(3x+2y)k\,

Which is equal to -y\sin xyi+(3y-4x)j-3k\, [Let this be equation 2]

{\frac  {\partial f}{\partial y}}={\frac  {\partial }{\partial y}}(\cos xy)i+{\frac  {\partial }{\partial y}}(3xy-2x^{2})j-{\frac  {\partial }{\partial y}}(3x+2y)k\,

Which is equal to -x\sin xyi+3xj-2k\, [Equation 3]

Partially differentiating equation 2,we get

{\frac  {\partial ^{2}f}{\partial x^{2}}}={\frac  {\partial }{\partial x}}[{\frac  {\partial f}{\partial x}}={\frac  {\partial }{\partial x}}[-y\sin xyi+(3y-4x)j-3kj]\,

This is equal to

{\frac  {\partial }{\partial x}}(-y\sin xy)i+{\frac  {\partial }{\partial x}}(3y-4x)j-{\frac  {\partial }{\partial x}}(3)k=-y^{2}\cos xyi-4j\,

Partially differentiating equation 3,we get

{\frac  {\partial ^{2}f}{\partial y^{2}}}={\frac  {\partial }{\partial y}}({\frac  {\partial f}{\partial y}}={\frac  {\partial }{\partial y}}(-x\sin xyi+3xj-2k)\,

{\frac  {\partial }{\partial y}}(-x\sin xy)i+{\frac  {\partial }{\partial y}}(3x)j-{\frac  {\partial }{\partial y}}(2)k=-x^{2}\cos xyi\,

And

{\frac  {\partial ^{2}f}{\partial x\partial y}}={\frac  {\partial }{\partial x}}({\frac  {\partial f}{\partial y}}\,

By equation 3

{\frac  {\partial }{\partial x}}[-x\sin xyi+3xj-2k]\,

{\frac  {\partial }{\partial x}}(-x\sin xy)i+{\frac  {\partial }{\partial x}}(3x)j-{\frac  {\partial }{\partial x}}(2)k=-(xy\cos xy+\sin xy)i+3j\,

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