Proofs Chapter 2: Gradient Formulas

Proof

The $i$-th component of the gradient is $[\nabla f]_i = \partial f / \partial x_i$. By the linearity of the partial derivative (the single-variable sum rule applied to each $x_i$),

\begin{equation}[\nabla(f+g)]_i = \dfrac{\partial (f+g)}{\partial x_i} = \dfrac{\partial f}{\partial x_i} + \dfrac{\partial g}{\partial x_i} = [\nabla f]_i + [\nabla g]_i \label{eq:13-1}\end{equation}

Since every component agrees, $\nabla(f+g) = \nabla f + \nabla g$.

Proof

A constant factor can be pulled out of a partial derivative, so for each component

\begin{equation}[\nabla(cf)]_i = \dfrac{\partial (cf)}{\partial x_i} = c\,\dfrac{\partial f}{\partial x_i} = c\,[\nabla f]_i \label{eq:14-1}\end{equation}

Hence $\nabla(cf) = c\,\nabla f$.

Proof

Apply the single-variable Leibniz rule to the partial derivative with respect to each $x_i$.

\begin{equation}[\nabla(fg)]_i = \dfrac{\partial (fg)}{\partial x_i} = f\,\dfrac{\partial g}{\partial x_i} + g\,\dfrac{\partial f}{\partial x_i} = [f\,\nabla g + g\,\nabla f]_i \label{eq:15-1}\end{equation}

Each component matches the corresponding component of $f\,\nabla g + g\,\nabla f$, which proves the formula.

Proof

Expand the $i$-th component of the gradient of $\mathbf{A}\cdot\mathbf{B} = \sum_j A_j B_j$ using the Leibniz rule.

\begin{equation}[\nabla(\mathbf{A}\cdot\mathbf{B})]_i = \dfrac{\partial}{\partial x_i}\!\left(\displaystyle\sum_j A_j B_j\right) = \displaystyle\sum_j \left( B_j\,\dfrac{\partial A_j}{\partial x_i} + A_j\,\dfrac{\partial B_j}{\partial x_i} \right) \label{eq:16-1}\end{equation}

Rewrite the right-hand side using the curl $\nabla \times \mathbf{A}$ and the directional derivative $(\mathbf{B}\cdot\nabla)\mathbf{A}$. Applying the BAC-CAB rule ($\mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})$) to $\mathbf{B} \times (\nabla \times \mathbf{A})$ in component form gives

\begin{equation}[\mathbf{B} \times (\nabla \times \mathbf{A})]_i = \displaystyle\sum_j B_j\,\dfrac{\partial A_j}{\partial x_i} - \displaystyle\sum_j B_j\,\dfrac{\partial A_i}{\partial x_j} \label{eq:16-2}\end{equation}

Swapping $\mathbf{A}$ and $\mathbf{B}$ similarly yields

\begin{equation}[\mathbf{A} \times (\nabla \times \mathbf{B})]_i = \displaystyle\sum_j A_j\,\dfrac{\partial B_j}{\partial x_i} - \displaystyle\sum_j A_j\,\dfrac{\partial B_i}{\partial x_j} \label{eq:16-3}\end{equation}

The second sum, $\sum_j B_j\,\partial A_i / \partial x_j$, is exactly the $i$-th component of $(\mathbf{B}\cdot\nabla)\mathbf{A}$. Solving $\eqref{eq:16-2}$ for the first sum,

\begin{equation}\displaystyle\sum_j B_j\,\dfrac{\partial A_j}{\partial x_i} = [\mathbf{B} \times (\nabla \times \mathbf{A})]_i + \displaystyle\sum_j B_j\,\dfrac{\partial A_i}{\partial x_j} = [\mathbf{B} \times (\nabla \times \mathbf{A}) + (\mathbf{B} \cdot \nabla)\mathbf{A}]_i \label{eq:16-4}\end{equation}

By the same argument,

\begin{equation}\displaystyle\sum_j A_j\,\dfrac{\partial B_j}{\partial x_i} = [\mathbf{A} \times (\nabla \times \mathbf{B}) + (\mathbf{A} \cdot \nabla)\mathbf{B}]_i \label{eq:16-5}\end{equation}

Substituting $\eqref{eq:16-4}$ and $\eqref{eq:16-5}$ into $\eqref{eq:16-1}$ shows that for every component

\begin{equation}[\nabla(\mathbf{A}\cdot\mathbf{B})]_i = [(\mathbf{B} \cdot \nabla)\mathbf{A} + (\mathbf{A} \cdot \nabla)\mathbf{B} + \mathbf{B} \times (\nabla \times \mathbf{A}) + \mathbf{A} \times (\nabla \times \mathbf{B})]_i \label{eq:16-6}\end{equation}

This establishes (16).

Proof

Apply the single-variable chain rule to each partial derivative:

\begin{equation}[\nabla f(g)]_i = \dfrac{\partial f(g)}{\partial x_i} = f'(g)\,\dfrac{\partial g}{\partial x_i} = [f'(g)\,\nabla g]_i \label{eq:17-1}\end{equation}

Since this holds for every $i$, we obtain $\nabla f(g) = f'(g)\,\nabla g$.

Proof

Apply (17) with the single-variable function $f(g) = g^n$. The single-variable power rule gives $f'(g) = n\,g^{n-1}$, hence by (17)

\begin{equation}\nabla f^n = n\,f^{n-1}\,\nabla f \label{eq:18-1}\end{equation}

which is an immediate special case of (17).

Proof

First, compute the gradient of $1/g$. Applying (17) with $f(g) = 1/g$ and $f'(g) = -1/g^2$,

\begin{equation}\nabla\!\left(\dfrac{1}{g}\right) = -\dfrac{1}{g^2}\,\nabla g \label{eq:19-1}\end{equation}

Now write $f/g = f \cdot (1/g)$ and combine the product rule (15) with $\eqref{eq:19-1}$:

\begin{equation}\nabla\!\left(\dfrac{f}{g}\right) = \nabla\!\left(f \cdot \dfrac{1}{g}\right) = \dfrac{1}{g}\,\nabla f + f\,\nabla\!\left(\dfrac{1}{g}\right) = \dfrac{1}{g}\,\nabla f - \dfrac{f}{g^2}\,\nabla g \label{eq:19-2}\end{equation}

Putting everything over the common denominator $g^2$,

\begin{equation}\nabla\!\left(\dfrac{f}{g}\right) = \dfrac{g\,\nabla f - f\,\nabla g}{g^2} \label{eq:19-3}\end{equation}

which proves the quotient rule.

Proof

Set $\mathbf{s} = \mathbf{r} - \mathbf{r}'$. Since $\mathbf{r}'$ is a fixed point, it is treated as a constant. The squared distance is

\begin{equation}\mathbf{s} = \mathbf{r} - \mathbf{r}', \qquad |\mathbf{s}|^2 = \displaystyle\sum_i (x_i - x'_i)^2 \label{eq:20-1}\end{equation}

Differentiating both sides with respect to $x_i$,

\begin{equation}2\,|\mathbf{s}|\,\dfrac{\partial |\mathbf{s}|}{\partial x_i} = 2(x_i - x'_i) \quad \Longrightarrow \quad \dfrac{\partial |\mathbf{s}|}{\partial x_i} = \dfrac{x_i - x'_i}{|\mathbf{s}|} \label{eq:20-2}\end{equation}

Rewriting in vector form,

\begin{equation}\nabla |\mathbf{s}| = \dfrac{\mathbf{s}}{|\mathbf{s}|} = \dfrac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|} \label{eq:20-3}\end{equation}

Now apply (17) with $f(g) = 1/g$ and $g = |\mathbf{s}|$, so that $f'(g) = -1/g^2$:

\begin{equation}\nabla\!\left(\dfrac{1}{|\mathbf{s}|}\right) = -\dfrac{1}{|\mathbf{s}|^2}\,\nabla |\mathbf{s}| = -\dfrac{1}{|\mathbf{s}|^2} \cdot \dfrac{\mathbf{s}}{|\mathbf{s}|} = -\dfrac{\mathbf{s}}{|\mathbf{s}|^3} \label{eq:20-4}\end{equation}

Substituting back $\mathbf{s} = \mathbf{r} - \mathbf{r}'$,

\begin{equation}\nabla\!\left(\dfrac{1}{|\mathbf{r} - \mathbf{r}'|}\right) = -\dfrac{\mathbf{r} - \mathbf{r}'}{|\mathbf{r} - \mathbf{r}'|^3} \label{eq:20-5}\end{equation}

This is the central formula used to derive the electric field $\mathbf{E} = -\nabla\phi$ from the electrostatic potential $1/(4\pi\varepsilon_0 |\mathbf{r}-\mathbf{r}'|)$ and is the direct mathematical expression of Coulomb's law.