VC3.41

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We shall use the identity {\mathrm  {grad}}(uv)=u{\mathrm  {grad}}v+v{\mathrm  {grad}}u\, --(1)

Now,{\mathrm  {grad}}u=i{\frac  {\partial }{\partial x}}(e^{{2x}}+x^{2}z)+j{\frac  {\partial }{\partial y}}(e^{{2x}}+x^{2}z)+k{\frac  {\partial }{\partial z}}(e^{{2x}}+x^{2}z)=i(2e^{{2x}}+2xz)+j(0)+k(x^{2})\, --(2)

and {\mathrm  {grad}}v=i{\frac  {\partial }{\partial x}}(2z^{2}y-xy^{2})+j{\frac  {\partial }{\partial y}}(2z^{2}y-xy^{2})+k{\frac  {\partial }{\partial z}}(2z^{2}y-xy^{2})=i(-y^{2})+j(2z^{2}-2xy)+k(4zy)\, --(3)

Using (2) and (3),(1) becomes

{\mathrm  {grad}}(uv)=(e^{{2x}}+x^{2}z)[-y^{2}i+(2z^{2}-2xy)j+4zyk]+(2z^{2}y-xy^{2})[(2e^{{2x}}+2xz)i+x^{2}k]\,

Putting x=1,y=0,z=2 in the above equation,we get

{\mathrm  {grad}}(uv)\, at(1,0,2)=(e^{2}+2)[8j]+0=8(2+e^{2})j\,

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