17. Divergence, Curl and Potentials

d. Differential Identities

1. First Order Differential Identities

These first order differential identities are much less important than the second order identities on the next page.

We will discuss these identities for scalar and vector fields in \(\mathbb R^3\) but they hold in any dimension unless they involve a cross product or a curl. In some of the proofs, we will only write out the \(x\)-component. The other components are analogous. Throughout these identities, \(f\) and \(g\) are scalar fields (functions), \(\vec F\) and \(\vec G\) are vector fields and \(h\) is a simple function of \(1\) variable. These first order identities generalize the product rule and the chain rule to the del operator, specifically to the gradient, divergence and curl. The only ones you need to remember are the first product rules under gradient, divergence and curl.

The Product Rule

• Gradient:

  \[ \vec\nabla(fg) =f\vec\nabla g+g\vec\nabla f \] Proof

  We start with the left side (and only show the \(x\) component): \[\begin{aligned} LHS &=\vec\nabla(f\,g) \\ &=\left\langle \partial_x(f\,g),\cdots,\cdots\right\rangle \\ &=\left\langle (f(\partial_x g)+g(\partial_x f),\cdots, \cdots\right\rangle \\ &=f\left\langle \partial_x g,\cdots, \cdots\right\rangle +g\left\langle \partial_x f,\cdots, \cdots\right\rangle \\ &=f\vec\nabla g+g\vec\nabla f =RHS \end{aligned}\]

  \[ \vec\nabla(\vec F\cdot\vec G) =(\vec F\cdot\vec\nabla)\vec G +(\vec G\cdot\vec\nabla)\vec F +\vec F\times(\vec\nabla\times\vec G) +\vec G\times(\vec\nabla\times\vec F) \] Proof

  We start with the right side (and only show the \(x\) component): \[\begin{aligned} RHS &=(\vec F\cdot\vec\nabla)\vec G +(\vec G\cdot\vec\nabla)\vec F +\vec F\times(\vec\nabla\times\vec G) +\vec G\times(\vec\nabla\times\vec F) \\ &=\left\langle F_1 \partial_x G_1 +F_2 \partial_y G_1 +F_3 \partial_z G_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_1 \partial_x F_1 +G_2 \partial_y F_1 +G_3 \partial_z F_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle F_2(\vec\nabla\times\vec G)_3-F_3(\vec\nabla\times\vec G)_2, \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_2(\vec\nabla\times\vec F)_3-G_3(\vec\nabla\times\vec F)_2, \cdots,\cdots\right\rangle \\ &=\left\langle F_1 \partial_x G_1 +F_2 \partial_y G_1 +F_3 \partial_z G_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_1 \partial_x F_1 +G_2 \partial_y F_1 +G_3 \partial_z F_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle F_2(\partial_x G_2-\partial_y G_1) -F_3(\partial_z G_1-\partial_x G_3), \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_2(\partial_x F_2-\partial_y F_1) -G_3(\partial_z F_1-\partial_x F_3), \cdots,\cdots\right\rangle \end{aligned}\] The terms \(F_2 \partial_y G_1\), \(F_3 \partial_z G_1\), \(G_2 \partial_y F_1\) and \(G_3 \partial_z F_1\) cancel leaving: \[\begin{aligned} RHS &=\left\langle F_1 \partial_x G_1,\cdots,\cdots\right\rangle \\ &\quad+\left\langle G_1 \partial_x F_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle F_2(\partial_x G_2)-F_3(-\partial_x G_3), \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_2(\partial_x F_2)-G_3(-\partial_x F_3), \cdots,\cdots\right\rangle \\ &=\left\langle F_1 \partial_x G_1 +G_1 \partial_x F_1 +F_2 \partial_x G_2 +G_2 \partial_x F_2\right. \\ &\quad\left.+\,F_3 \partial_x G_3 +G_3 \partial_x F_3, \cdots,\cdots\right\rangle \end{aligned}\] Finally, we use the product rule in reverse: \[\begin{aligned} RHS &=\left\langle \partial_x(F_1 G_1 +F_2 G_2 +F_3 G_3), \cdots,\cdots\right\rangle \\ &=\left\langle \partial_x(\vec F\cdot\vec G), \cdots,\cdots\right\rangle \\ &=\vec\nabla(\vec F\cdot\vec G)=LHS\\ \end{aligned}\]

  Here, the operator \((\vec F\cdot\vec\nabla)\) is the product of \(\vec F\) and \(\vec\nabla\) without differentiating: \[ (\vec F\cdot\vec\nabla) =F_1 \partial_x +F_2 \partial_y +F_3 \partial_z \] Then \((\vec F\cdot\vec\nabla)\vec G\) means to apply this operator to each component of \(\vec G\).

If \(f=x^2yz^3\) and \(g=x^2y^3z^2\), compute \(\vec\nabla(fg)\) in two ways.

1. Find \(f\vec\nabla g\) and \(g\vec\nabla f\) and add them to get \(\vec\nabla(fg)\).

   \(\begin{aligned} f\vec\nabla g &=\left\langle 2x^3y^4z^5,3x^4y^3z^5,2x^4y^4z^4\right\rangle \\ g\vec\nabla f &=\left\langle 2x^3y^4z^5,x^4y^3z^5,3x^4y^4z^4\right\rangle \\ \vec\nabla(fg) &=\left\langle 4x^3y^4z^5,4x^4y^3z^5,5x^4y^4z^4\right\rangle \end{aligned}\)

   \[\begin{aligned} f\vec\nabla g &=x^2yz^3\left\langle 2xy^3z^2,3x^2y^2z^2,2x^2y^3z\right\rangle \\ &=\left\langle 2x^3y^4z^5,3x^4y^3z^5,2x^4y^4z^4\right\rangle \\[6pt] g\vec\nabla f &=x^2y^3z^2\left\langle 2xyz^3,x^2z^3,3x^2yz^2\right\rangle \\ &=\left\langle 2x^3y^4z^5,x^4y^3z^5,3x^4y^4z^4\right\rangle \\[6pt] \vec\nabla(fg) &=f\vec\nabla g+g\vec\nabla f \\ &=\left\langle 4x^3y^4z^5,4x^4y^3z^5,5x^4y^4z^4\right\rangle \end{aligned}\]

2. Find \(fg\) and then \(\vec\nabla(fg)\).

   \(\begin{aligned} fg&=x^4y^4z^5 \\ \vec\nabla(fg) &=\left\langle 4x^3y^4z^5,4x^4y^3z^5,5x^4y^4z^4\right\rangle \end{aligned}\)

   \[\begin{aligned} fg&=x^4y^4z^5 \\[6pt] \vec\nabla(fg) &=\left\langle 4x^3y^4z^5,4x^4y^3z^5,5x^4y^4z^4\right\rangle \end{aligned}\]

• Divergence:

  \[ \vec\nabla\cdot(f\vec G) =(\vec\nabla f)\cdot\vec G +f\vec\nabla\cdot\vec G \] Proof

  We start with the left side: \[\begin{aligned} LHS &=\vec\nabla\cdot(f\vec G) \\ &=\vec\nabla\cdot\left\langle fG_1,fG_2,fG_3\right\rangle \\ &=\partial_x(fG_1)+\partial_y(fG_2)+\partial_z(fG_3) \end{aligned}\] We apply the product rule for ordinary functions: \[\begin{aligned} LHS &=(\partial_xf)G_1 +f\partial_xG_1 +(\partial_yf)G_2 +f\partial_yG_2 \\ &\quad+(\partial_zf)G_3 +f\partial_zG_3 \\ &=(\vec\nabla f)\cdot\vec G +f\vec\nabla\cdot\vec G =RHS \end{aligned}\]

  \[ \vec\nabla\cdot(\vec F\times\vec G) =\vec\nabla\times\vec F\cdot\vec G-\vec F\cdot\vec\nabla\times\vec G \] Proof

  We start with the left side: \[\begin{aligned} LHS &=\vec\nabla\cdot(\vec F\times\vec G) \\ &=\partial_x(F_2G_3-F_3G_2) +\partial_y(F_3G_1-F_1G_3) \\ &\quad+\partial_z(F_1G_2-F_2G_1) \end{aligned}\] We apply the product rule for ordinary functions: \[\begin{aligned} LHS &=(\partial_xF_2)G_3+F_2(\partial_xG_3) -(\partial_xF_3)G_2-F_3(\partial_xG_2) \\ &\quad+(\partial_yF_3)G_1+F_3(\partial_yG_1) -(\partial_yF_1)G_3-F_1(\partial_yG_3) \\ &\quad+(\partial_zF_1)G_2+F_1(\partial_zG_2) -(\partial_zF_2)G_1-F_2(\partial_zG_1) \\ &=(\partial_yF_3-\partial_zF_2)G_1 +(\partial_zF_1-\partial_xF_3)G_2 +(\partial_xF_2-\partial_yF_1)G_3 \\ &\quad-F_1(\partial_yG_3-\partial_zG_2) -F_2(\partial_zG_1-\partial_xG_3) -F_3(\partial_xG_2-\partial_yG_1) \\ &=\vec\nabla\times\vec F\cdot\vec G-\vec F\cdot\vec\nabla\times\vec G =RHS \end{aligned}\]

  In the divergence of a cross product, notice that both sides flip signs when \(\vec F\) and \(\vec G\) are interchanged. This is a reason for the minus sign before the second term. It comes from interchanging the order of the \(\vec F\) and the \(\vec\nabla\). You can think of this as the minus sign that results when you interchange two vectors in a triple product.

If \(f=x^2yz^3\) and \(\vec G=\left\langle x^4,y^3,z^2\right\rangle\), compute \(\vec\nabla\cdot (f\vec G)\) in two ways.

1. Find \((\vec\nabla f)\cdot\vec G\) and \(f\vec\nabla\cdot\vec G\) and add them to get \(\vec\nabla\cdot(f\vec G)\).

   \(\begin{aligned} (\vec\nabla f)\cdot\vec G &=2x^5yz^3+x^2y^3z^3+3x^2yz^4 \\ f\vec\nabla\cdot\vec G &=4x^5yz^3+3x^2y^3z^3+2x^2yz^4 \\ \vec\nabla\cdot(f\vec G) &=6x^5yz^3+4x^2y^3z^3+5x^2yz^4 \end{aligned}\)

   \[\begin{aligned} (\vec\nabla f) &=\left\langle 2xyz^3,x^2z^3,3x^2yz^2\right\rangle \\ (\vec\nabla f)\cdot\vec G &=2x^5yz^3+x^2y^3z^3+3x^2yz^4 \\[6pt] \vec\nabla\cdot\vec G &=4x^3+3y^2+2z \\ f\vec\nabla\cdot\vec G &=4x^5yz^3+3x^2y^3z^3+2x^2yz^4 \\[6pt] \vec\nabla\cdot(f\vec G) &=(\vec\nabla f)\cdot\vec G +f\vec\nabla\cdot\vec G \\ &=6x^5yz^3+4x^2y^3z^3+5x^2yz^4 \end{aligned}\]

2. Find \(f\vec G\) and then \(\vec\nabla\cdot(f\vec G)\).

   \(\begin{aligned} f\vec G &=\left\langle x^6yz^3,x^2y^4z^3,x^2yz^5\right\rangle \\ \vec\nabla\cdot(f\vec G) &=6x^5yz^3+4x^2y^3z^3+5x^2yz^4 \end{aligned}\)

   \[\begin{aligned} f\vec G &=x^2yz^3\left\langle x^4,y^3,z^2\right\rangle \\ &=\left\langle x^6yz^3,x^2y^4z^3,x^2yz^5\right\rangle \\[6pt] \vec\nabla\cdot(f\vec G) &=6x^5yz^3+4x^2y^3z^3+5x^2yz^4 \end{aligned}\]

• Curl:

  \[ \vec\nabla\times(f\vec G) =(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \] Proof

  We start with the left side (and only show the \(x\) component): \[\begin{aligned} LHS &=\vec\nabla\times(f\vec G) \\ &=\left\langle \partial_y(fG_3)-\partial_z(fG_2), \cdots,\cdots\right\rangle \\ \end{aligned}\] We apply the product rule for ordinary functions: \[\begin{aligned} LHS &=\left\langle (\partial_y f)G_3+f\partial_y G_3 -(\partial_z f)G_2-f\partial_z G_2, \cdots,\cdots\right\rangle \\ &=(\vec\nabla f)\times\vec G +f(\vec\nabla\times\vec G) =RHS \end{aligned}\]

  \[ \vec\nabla\times(\vec F\times\vec G) =(\vec G\cdot\vec\nabla)\vec F -(\vec F\cdot\vec\nabla)\vec G +(\vec\nabla\cdot\vec G)\vec F -(\vec\nabla\cdot\vec F)\vec G \] Proof

  We start with the right side (and only show the \(x\) component): \[\begin{aligned} RHS &=(\vec G\cdot\vec\nabla)\vec F-(\vec F\cdot\vec\nabla)\vec G +(\vec\nabla\cdot\vec G)\vec F-(\vec\nabla\cdot\vec F)\vec G \\ &=\left\langle G_1(\partial_x F_1)+G_2(\partial_y F_1)+G_3(\partial_z F_1), \cdots,\cdots\right\rangle \\ &\quad-\left\langle F_1(\partial_x G_1)+F_2(\partial_y G_1)+F_3(\partial_z G_1), \cdots,\cdots\right\rangle \\ &\quad+\left\langle(\partial_x G_1)F_1 +(\partial_y G_2)F_1 +(\partial_z G_3)F_1, \cdots,\cdots\right\rangle \\ &\quad-\left\langle(\partial_x F_1)G_1 +(\partial_y F_2)G_1 +(\partial_z F_3)G_1, \cdots,\cdots\right\rangle \\ \end{aligned}\] The terms \(G_1(\partial_x F_1)\) and \(F_1(\partial_x G_1)\) cancel and we rearrange the terms leaving: \[\begin{aligned} RHS &=\left\langle G_2(\partial_y F_1)+(\partial_y G_2)F_1, \cdots,\cdots\right\rangle \\ &\quad-\left\langle F_2(\partial_y G_1)+(\partial_y F_2)G_1, \cdots,\cdots\right\rangle \\ &\quad+\left\langle G_3(\partial_z F_1)+(\partial_z G_3)F_1, \cdots,\cdots\right\rangle \\ &\quad-\left\langle F_3(\partial_z G_1)+(\partial_z F_3)G_1, \cdots,\cdots\right\rangle \\ \end{aligned}\] We now use the product rule in reverse: \[\begin{aligned} RHS &=\left\langle \partial_y(F_1 G_2),\cdots,\cdots\right\rangle -\left\langle \partial_y(F_2 G_1),\cdots,\cdots\right\rangle \\ &\quad+\left\langle \partial_z(F_1 G_3),\cdots,\cdots\right\rangle -\left\langle \partial_z(F_3 G_1),\cdots,\cdots\right\rangle \\ &=\left\langle \partial_y(F_1 G_2-F_2 G_1),\cdots,\cdots\right\rangle \\ &\quad-\left\langle \partial_z(F_3 G_1-F_1 G_3),\cdots,\cdots\right\rangle \\ &=\left\langle \partial_y(\vec F\times\vec G)_3 -\partial_z(\vec F\times\vec G)_2,\cdots,\cdots\right\rangle \\ &=\vec\nabla\times(\vec F\times\vec G) =LHS \end{aligned}\]

  In the curl of a cross product, notice that both sides flip signs when \(\vec F\) and \(\vec G\) are interchanged.

If \(f=x^2yz^3\) and \(\vec G=\left\langle xy,yz,zx\right\rangle\), compute \(\vec\nabla\times (f\vec G)\) in two ways.

1. Find \((\vec\nabla f)\times\vec G\) and \(f\vec\nabla\times\vec G\) and add them to get \(\vec\nabla\times(f\vec G)\).

   \(\begin{aligned} (\vec\nabla f)\times\vec G &=\left\langle x^3z^4-3x^2y^2z^3,3x^3y^2z^2-2x^2yz^4, 2xy^2z^4-x^3yz^3\right\rangle \\ f\vec\nabla\times\vec G &=\left\langle -x^2y^2z^3,-x^2yz^4,-x^3yz^3\right\rangle \\ \vec\nabla\times(f\vec G) &=\left\langle x^3z^4-4x^2y^2z^3, 3x^3y^2z^2-3x^2yz^4, 2xy^2z^4-2x^3yz^3\right\rangle \end{aligned}\)

   \[\begin{aligned} \vec\nabla f &=\left\langle 2xyz^3,x^2z^3,3x^2yz^2\right\rangle \\[6pt] (\vec\nabla f)\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2xyz^3 & x^2z^3 & 3x^2yz^2 \\ xy & yz & zx \end{vmatrix} \\ &=\hat\imath(x^3z^4-3x^2y^2z^3)-\hat\jmath(2x^2yz^4-3x^3y^2z^2) \\ &\quad+\hat k(2xy^2z^4-x^3yz^3) \\ &=\left\langle x^3z^4-3x^2y^2z^3,3x^3y^2z^2-2x^2yz^4, 2xy^2z^4-x^3yz^3\right\rangle \\[6pt] \vec\nabla\times\vec G &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ xy & yz & zx \end{vmatrix} \\ &=\hat\imath(0-y)-\hat\jmath(z-0) +\hat k(0-x) \\ &=\left\langle -y,-z,-x\right\rangle \\[6pt] f\vec\nabla\times\vec G &=\left\langle -x^2y^2z^3,-x^2yz^4,-x^3yz^3\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=(\vec\nabla f)\times\vec G +f\vec\nabla\times\vec G \\ &=\left\langle x^3z^4-4x^2y^2z^3, 3x^3y^2z^2-3x^2yz^4, 2xy^2z^4-2x^3yz^3\right\rangle \end{aligned}\]

2. Find \(f\vec G\) and then \(\vec\nabla\times(f\vec G)\).

   \(\begin{aligned} f\vec G &=\left\langle x^3y^2z^3,x^2y^2z^4,x^3yz^4\right\rangle \\ \vec\nabla\times(f\vec G) &=\left\langle x^3z^4-4x^2y^2z^3, 3x^3y^2z^2-3x^2yz^4, 2xy^2z^4-2x^3yz^3\right\rangle \end{aligned}\)

   \[\begin{aligned} f\vec G &=\left\langle x^3y^2z^3,x^2y^2z^4,x^3yz^4\right\rangle \\[6pt] \vec\nabla\times(f\vec G) &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ x^3y^2z^3 & x^2y^2z^4 & x^3yz^4 \end{vmatrix} \\ &=\hat\imath(x^3z^4-4x^2y^2z^3)-\hat\jmath(3x^2yz^4-3x^3y^2z^2) +\hat k(2xy^2z^4-2x^3yz^3) \\ &=\left\langle x^3z^4-4x^2y^2z^3, 3x^3y^2z^2-3x^2yz^4, 2xy^2z^4-2x^3yz^3\right\rangle \end{aligned}\]

The Chain Rule

• Gradient:

  \[ \vec\nabla(f(u)) =f'(u)\vec\nabla u \] Proof

  We start from the left side: \[ LHS =\vec\nabla(f(u)) =\left\langle \partial_x(f(u)), \partial_y(f(u)), \partial_z(f(u))\right\rangle \] We use the chain rule for a partial derivative of an ordinary composition of functions: \[\begin{aligned} LHS &=\left\langle f'(u)\partial_x u,f'(u)\partial_y u,f'(u)\partial_z u\right\rangle \\ &=f'(u)\vec\nabla u =RHS \end{aligned}\]

Compute \(\vec\nabla(\sin(x^2yz^3))\) in two ways.

1. Let \(u=\sin(w)\) and \(f=x^2yz^3\). Find \(u'(f)\), \(\vec\nabla f\) and \(u'(f)\vec\nabla f\).

   \(u'(f)\vec\nabla f =\left\langle \cos(x^2yz^3)2xyz^3,\cos(x^2yz^3)x^2z^3, \cos(x^2yz^3)3x^2yz^2\right\rangle\)

   If \(u=\sin(w)\) and \(f=x^2yz^3\), then \(u'=\cos(w)\) and \(u'(f)=\cos(x^2yz^3)\). Further, \(\vec\nabla f=\left\langle 2xyz^3,x^2z^3,3x^2yz^2\right\rangle\). So \[ u'(f)\vec\nabla f =\left\langle \cos(x^2yz^3)2xyz^3,\cos(x^2yz^3)x^2z^3, \cos(x^2yz^3)3x^2yz^2\right\rangle \]

2. Compute \(\vec\nabla(\sin(x^2yz^3))\) by taking the \(x\), \(y\) and \(z\) derivatives of \(\sin(x^2yz^3)\).

   \(\vec\nabla \sin(x^2yz^3) =\left\langle \cos(x^2yz^3)2xyz^3,\cos(x^2yz^3)x^2z^3, \cos(x^2yz^3)3x^2yz^2\right\rangle\)

   We compute the partial derivatives of \(\sin(x^2yz^3)\): \[\begin{aligned} \partial_x\sin(x^2yz^3) &=\cos(x^2yz^3)2xyz^3 \\ \partial_y\sin(x^2yz^3) &=\cos(x^2yz^3)x^2z^3 \\ \partial_z\sin(x^2yz^3) &=\cos(x^2yz^3)3x^2yz^2 \\ \end{aligned}\] So \[ \vec\nabla \sin(x^2yz^3) =\left\langle \cos(x^2yz^3)2xyz^3,\cos(x^2yz^3)x^2z^3, \cos(x^2yz^3)3x^2yz^2\right\rangle \]