Find $\nabla (\vert \overrightarrow r \rvert^2)$

Question

Extra background info:

$\overrightarrow r$ is the vector from the source point ($\overrightarrow a$) to the field point ($\overrightarrow b$).

$\overrightarrow a=(x',y',z')$

$\overrightarrow b=(x,y,z)$

Screenshot of the problem in the book

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I'm having trouble with just one part. I don't know how to take the derivative of each component.

Given that $\vert \overrightarrow r \rvert=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}$, $\vert \overrightarrow r \rvert ^2$ = $(x-x')^2+(y-y')^2+(z-z')^2$.

$\nabla (\vert \overrightarrow r \rvert ^2)= \hat x \frac{\partial r^2}{\partial x}+\hat y \frac{\partial r^2}{\partial y}+\hat z \frac{\partial r^2}{\partial z}$

But here is where I'm stuck. When we have $x-x'$, how do I take the derivative of this? Do I only take the derivative the first $x$ because the denominator of the operator says $x$ and not $x'$?

Answer 1

Here you have \begin{align*} \frac{\partial r^2}{\partial x} &= 2(x-x') \end{align*}

Therefore

\begin{align*} \nabla (r^2) &= 2 \,[ \ \hat x (x-x')+\hat y (y-y')+\hat z(z-z') \ ]\\ &=2(\mathbf{b}-\mathbf{a}) \end{align*}

According to the screen shot of the problem you have $$ \nabla (r^2) = 2 \mathbf{r} $$