VC1.15

From Exampleproblems

Here $\phi f=xy^2 z(xzi-xyj+yz^2k)=x^2 y^2 z^2 i-x^2 y^3 z j+xy^3 z^3k\,$

Now $\frac{\partial }{\partial z}(\phi f)=\frac{\partial }{\partial z}(x^2 y^2 z^2)i-\frac{\partial }{\partial z}(x^2 y^3 z)j+\frac{\partial }{\partial z}(xy^3 z^3)k\,$ which is equal to

$2x^2 y^2 zi-x^2 y^3j+3xy^3 z^2k\,$

Now differentiating the above,we get

$\frac{\partial^2 (\phi f) }{\partial x \partial z}=\frac{\partial }{\partial x}[\frac{\partial (\phi f) }{\partial z}]\,$ which is equal to

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

= $4xy^2zi-2xy^3j+3y^3z^2k\,$

Again partially differentiating the above result,

$\frac{\partial^3 (\phi f)}{\partial x^2 \partial z}=\frac{\partial }{\partial x}[\frac{\partial^2 (\phi f)}{\partial x \partial z}]\,$

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

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

= $4y^2zi-2y^3j\,$

Therefore at A(2,-1,1),

$\frac{\partial^3 (\phi f)}{\partial x^2 \partial z}=4(-1)^2(1)i-2(-1)^3j=4i+2j\,$

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