VC1.13

From Exampleproblems

Given, $f=\cos xy i+(3xy-2x^2)j-(3x+2y)k\,$

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 $-y\sin xy i+(3y-4x)j-3k\,$ [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 $-x\sin xy i+3xj-2k\,$ [Equation 3]

Partially differentiating equation 2,we get

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

This is equal to

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

Partially differentiating equation 3,we get

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

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

And

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

By equation 3

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

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

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