Question

Derive the three quotient rules.

Short answer

The three quotient rules$$\begin{gathered} \nabla \left(\frac{f}{g}\right)=\frac{g\left(\nabla f\right)-f\left(\nabla g\right)}{g^2},\nabla \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)=\frac{g\left(\nabla \stackrel{\rightharpoonup }{A}\right)-\stackrel{\rightharpoonup }{A}\left(\nabla g\right)}{g^2} \\ \mathrm{and}\nabla \times \left(\frac{\stackrel{\rightharpoonup }{\mathrm{A}}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\nabla \times \stackrel{\rightharpoonup }{\mathrm{A}}\right)-\stackrel{\rightharpoonup }{\mathrm{A}}\times \left(\nabla \mathrm{g}\right)}{{\mathrm{g}}^2}\mathrm{have}\mathrm{been}\mathrm{derived}. \end{gathered}$$

Step-by-step solution

Derive first quiotinet rule

To compute an expression substitute the vectors and other required expression and then simplify

The first quiotient rule is$\nabla \left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{\mathrm{g}\left(\nabla \mathrm{f}\right)-\mathrm{f}\left(\nabla \mathrm{g}\right)}{{\mathrm{g}}^2},$The$\nabla$operator is defined aslocalid="1657516180871" $$\begin{gathered} \nabla =\left(\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{z}}\mathrm{k}\right) \\ \mathrm{Compute}\mathrm{the}\mathrm{value}\mathrm{of}\nabla \left(\frac{\mathrm{f}}{\mathrm{g}}\right),\mathrm{as}: \\ \nabla \left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\left(\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{z}}\mathrm{k}\right)\left(\frac{\mathrm{f}}{\mathrm{g}}\right) \\ =\mathrm{i}\frac{\partial }{\partial \mathrm{x}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)+\mathrm{j}\frac{\partial }{\partial \mathrm{y}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)\mathrm{k}+\frac{\partial }{\partial \mathrm{z}}\left(\frac{\mathrm{f}}{\mathrm{g}}\right) \\ =\left(\frac{\mathrm{g}{\displaystyle \frac{\partial \mathrm{f}}{\partial \mathrm{x}}}-\mathrm{f}\left({\displaystyle \frac{\partial \mathrm{g}}{\partial \mathrm{x}}}\right)}{{\mathrm{g}}^2}\right)+\mathrm{j}\left(\frac{\mathrm{g}{\displaystyle \frac{\partial \mathrm{f}}{\partial \mathrm{y}}}-\mathrm{f}\left({\displaystyle \frac{\partial \mathrm{g}}{\partial \mathrm{y}}}\right)}{{\mathrm{g}}^2}\right)+\mathrm{k}\left(\frac{\mathrm{g}{\displaystyle \frac{\partial \mathrm{f}}{\partial \mathrm{z}}}-\mathrm{f}\left({\displaystyle \frac{\partial \mathrm{g}}{\partial \mathrm{z}}}\right)}{{\mathrm{g}}^2}\right) \\ =\frac{1}{\mathrm{g}}\left(\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial \mathrm{f}}{\partial \mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\frac{\partial \mathrm{g}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{g}}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial \mathrm{g}}{\partial \mathrm{z}}\mathrm{k}\right) \end{gathered}$$

$\mathrm{Substitiute}\nabla \mathrm{for}\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{z}}\mathrm{k}\mathrm{into}\mathrm{equation}\left(1\right),\mathrm{as}:$

$$\begin{gathered} \nabla \left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\frac{1}{g}\left(\frac{\partial f}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial f}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial f}{\partial \mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\frac{\partial g}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial g}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial g}{\partial \mathrm{z}}\mathrm{k}\right) \\ =\frac{1}{g}\left(\nabla f\right)-\frac{1}{g^2}\left(\nabla g\right) \\ =\frac{g\left(\nabla f\right)-f\left(\nabla g\right)}{g^2} \end{gathered}$$

Thus the first quoitient rule is proven.

Derive second quiotient rule 

To compute an expression substitute the vectors and other required expression and then simplify

The second quiotient rule is $\nabla \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)=\frac{g\left(\nabla \stackrel{\rightharpoonup }{A}\right)-\stackrel{\rightharpoonup }{A}\left(\nabla g\right)}{g^2}$. Let the vector $\stackrel{\rightharpoonup }{A}$is defined as The operator is defined as $$\begin{gathered} \stackrel{\rightharpoonup }{A}=A_{x}i+A_{y}j+A_{z}k\mathrm{The}\nabla \mathrm{operator}\mathrm{is}\mathrm{defined}\mathrm{as} \\ \nabla =\frac{\partial }{\partial \mathrm{x}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{z}}\mathrm{k}. \end{gathered}$$

$$\begin{gathered} \mathrm{Compute}\mathrm{the}\mathrm{value}\mathrm{of}\nabla \left(\frac{\stackrel{\rightharpoonup }{\mathrm{A}}}{\mathrm{g}}\right),\mathrm{as}: \\ \nabla \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right)\left(\frac{A_{x}i+A_{y}i+A_{z}k}{g}\right) \\ =i\frac{\partial }{\partial x}\left(\frac{A_x}{g}\right)+j\frac{\partial }{\partial y}\left(\frac{A_y}{g}\right)+j\frac{\partial }{\partial z}\left(\frac{A_z}{g}\right) \\ =\frac{g{\displaystyle \frac{\partial A_x}{\partial x}}-A_x\left({\displaystyle \frac{\partial g}{\partial x}}\right)}{g^2}+\frac{g\frac{\partial A_y}{\partial y}-A_y\left(\frac{\partial g}{\partial y}\right)}{g^2}+\frac{g\frac{\partial A_z}{\partial z}-A_z\left(\frac{\partial g}{\partial z}\right)}{g^2} \\ =\frac{1}{g}\left(\frac{\partial A_x}{\partial x}+\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)-\frac{f}{g^2}\left(\frac{\partial g}{\partial x}+A_y\frac{\partial g}{\partial y}+A_Z\frac{\partial g}{\partial z}\right)......\left(1\right) \end{gathered}$$

The dot product of $\nabla$and vector $\stackrel{\rightharpoonup }{A}$is obtained as ,

$$\begin{gathered} \nabla .\stackrel{\rightharpoonup }{A}=\left(\frac{\partial }{\partial X}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial Z}k\right)\left(A_{x}i+A_{y}j+A_{z}k\right) \\ =\frac{A_x}{\partial X}i+\frac{\partial A_y}{\partial y}j+\frac{\partial A_z}{\partial Z}k \\ \mathrm{The}\mathrm{dot}\mathrm{product}\mathrm{of}\nabla \mathrm{g}\mathrm{and}\mathrm{vector}\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{is}\mathrm{obtained}\mathrm{as}, \\ \stackrel{\rightharpoonup }{\mathrm{A}}.\left(\nabla \mathrm{g}\right)=\left({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k}\right)\left(\frac{\partial }{\partial \mathrm{X}}\mathrm{i}+\frac{\partial }{\partial \mathrm{y}}\mathrm{j}+\frac{\partial }{\partial \mathrm{Z}}\mathrm{k}\right)\mathrm{g} \\ =\left({\mathrm{A}}_{\mathrm{x}}\mathrm{i}+{\mathrm{A}}_{\mathrm{y}}\mathrm{j}+{\mathrm{A}}_{\mathrm{z}}\mathrm{k}\right)\left(\frac{\partial \mathrm{g}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial \mathrm{g}}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial \mathrm{g}}{\partial \mathrm{z}}\mathrm{k}\right) \\ ={\mathrm{A}}_{\mathrm{x}}\frac{\partial \mathrm{g}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial \mathrm{g}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{x}}\frac{\partial \mathrm{g}}{\partial \mathrm{z}} \end{gathered}$$

$$\begin{gathered} \mathrm{Substitiute}\stackrel{\rightharpoonup }{\mathrm{A}}-\left(\nabla \mathrm{g}\right)\mathrm{for}{\mathrm{A}}_{\mathrm{x}}\frac{\partial \mathrm{g}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial \mathrm{g}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial \mathrm{g}}{\partial \mathrm{y}}\mathrm{and}\nabla -\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{for}\frac{{\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{z}} \\ \mathrm{into}\mathrm{equation}\left(1\right),\mathrm{as}: \\ \nabla \left(\frac{\stackrel{\rightharpoonup }{\mathrm{A}}}{\mathrm{g}}\right)=\frac{1}{\mathrm{g}}\left(\frac{\partial {\mathrm{A}}_{\mathrm{x}}}{\partial \mathrm{x}}\mathrm{i}+\frac{\partial {\mathrm{A}}_{\mathrm{y}}}{\partial \mathrm{y}}\mathrm{j}+\frac{\partial {\mathrm{A}}_{\mathrm{z}}}{\partial \mathrm{z}}\mathrm{k}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left({\mathrm{A}}_{\mathrm{x}}\frac{\partial \mathrm{g}}{\partial \mathrm{x}}+{\mathrm{A}}_{\mathrm{y}}\frac{\partial \mathrm{g}}{\partial \mathrm{y}}+{\mathrm{A}}_{\mathrm{z}}\frac{\partial \mathrm{g}}{\partial \mathrm{z}}\right) \\ =\frac{1}{\mathrm{g}}\left(\nabla .\stackrel{\rightharpoonup }{\mathrm{A}}\right)-\frac{\mathrm{f}}{{\mathrm{g}}^2}\left(\stackrel{\rightharpoonup }{\mathrm{A}}.\nabla \mathrm{g}\right) \\ =\frac{\mathrm{g}\left(\nabla .\stackrel{\rightharpoonup }{\mathrm{A}}\right)-\stackrel{\rightharpoonup }{\mathrm{A}}\left(\nabla \mathrm{g}\right)}{{\mathrm{g}}^2} \end{gathered}$$

Thus the second quoitient rule is proven.

Derive third quiotient rule 

To compute an expression substitute the vectors and other required expression and then simplify

The second quiotient rule is $\nabla \times \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)=\frac{g\left(\nabla \times \stackrel{\rightharpoonup }{A}\right)-\stackrel{\rightharpoonup }{A}\left(\nabla g\right)}{g^2}$. Let the vector $\stackrel{\rightharpoonup }{A}$is defined as $\stackrel{\rightharpoonup }{A}=A_{x}i+A_{x}j+A_{x}k$The operator is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$.

Compute the value of $\nabla \times \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)$, as:

$$\begin{gathered} \nabla \left(\frac{\stackrel{\rightharpoonup }{A}}{g}\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \frac{A_x}{g}& \frac{A_y}{g}& \frac{A_z}{g}\end{array}\right| \\ =i\left(\frac{\partial }{\partial y}\left(\frac{A_z}{g}\right)-\frac{\partial }{\partial z}\left(\frac{A_y}{g}\right)\right)+j\left(\frac{\partial }{\partial z}\left(\frac{A_x}{g}\right)-\frac{\partial }{\partial x}\left(\frac{A_z}{g}\right)\right)+k\left(\frac{\partial }{\partial x}\left(\frac{A_y}{g}\right)-\frac{\partial }{\partial y}\left(\frac{A_x}{g}\right)\right) \\ =\frac{1}{g}\left(i\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)+j\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)+k\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\right) \\ =\frac{g\left(\nabla \times \stackrel{\rightharpoonup }{A}\right)-\stackrel{\rightharpoonup }{A}\times \left(\nabla g\right)}{g^2} \end{gathered}$$

Thus the third quoitient rule is proven.