VC1.15

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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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