VC1.14

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

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

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

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

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

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

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

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

And

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

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

$4xi+(e^{xy}+xye^{xy}-\cos x)j-2x\sin y k\,$ [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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