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Given,f=(2x^{2}y-x^{4})i+(e^{{xy}}-y\sin x)j+x^{2}\cos yk\, [Equation 1]

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

Which is equal to

(4xy-4x^{3})i+(ye^{{xy}}-y\cos x)j+2x\cos yk\, [Equation 2]

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

2x^{2}i+(xe^{{xy}}-\sin x)j-x^{2}\sin yk\,

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

{\frac  {\partial }{\partial x}}(2x^{2}i+(xe^{{xy}}-\sin x)j-x^{2}\sin yk)\,

{\frac  {\partial }{\partial x}}(2x^{2})i+{\frac  {\partial }{\partial x}}(xe^{{xy}}-\sin x)j-{\frac  {\partial }{\partial x}}(x^{2}\sin y)k\, which is equal to

4xi+(xye^{{xy}}+e^{{xy}}-\cos x)j-2x\sin yk\, [Equation 3]


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

Using equation 2,

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

Which is equal to

4xi+(e^{{xy}}+xye^{{xy}}-\cos x)j-2x\sin yk\, [Equation 4]

From equations 3 and 4

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

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