Trying to understand the Nabla Operator

Question

I'm trying to wrap my head around the following line done in my physics textbook:

$\vec\nabla f(r) = \begin{pmatrix} f'(r) \frac{\partial r}{\partial x}\\ f'(r) \frac{\partial r}{\partial y}\\ f'(r) \frac{\partial r}{\partial z} \end{pmatrix}$

Where $r$ represents the distance between the origin and some object (int three dimensional space). Therefore I could write it as $r = \sqrt{x^2 + y^2 + z^2}$.

As far as I know $\vec\nabla$ represents a vector, which components are the partial derivatives of a given function and $\vec\nabla f(r)$ should be equal to $\begin{pmatrix}\frac{\partial f}{\partial r} \end{pmatrix}$ or simply $f'(r)$ since $f(r)$ only depends on $r$. I could write $f(r)$ as a function of $x,y,z$ and

$\vec\nabla f(x,y,z)$ would be equal to $\begin{pmatrix} \frac{\partial r}{\partial x}\\ \frac{\partial r}{\partial y}\\ \frac{\partial r}{\partial z} \end{pmatrix}$

How do I get the result of my textbook and what part did I misunderstand?

Answer 1

The only thing you are misunderstanding is that saying "I could write it as $r=\sqrt{x^2+y^2+z^2}$," you are essentially saying that $r = r(x,y,z)$. Therefore, what you are actually looking for is:

$$\nabla f(r) = \nabla f(r(x,y,z))=\pmatrix{\frac{\partial f(r(x,y,z))}{\partial x}\\ \frac{\partial f(r(x,y,z))}{\partial y}\\ \frac{\partial f(r(x,y,z))}{\partial z}\\}$$

And finally, applying the chain rule to each of those elements of the vector, you reach the result of your textbook.

Answer 2

It is a consequence of the derivative of function.

You have $f(r)$, with $r=\sqrt{x^2+y^2+z^2}$

Then $\dfrac{\partial f(r)}{\partial x}=\dfrac{\partial f(r)}{\partial r}\times\dfrac{\partial r}{\partial x}$

Answer 3

You are almost correct. We can write

$$\nabla f(r)=\hat r f'(r) =\frac{\vec r}{r}f'(r) \tag 1$$

where $\hat r=\vec r/r$ is the radial unit vector, $r=\sqrt{x^2+y^2+z^2}$ is the radial variable, and $\vec r=\hat rr=\hat xx+\hat yy+\hat zz$ is the position vector.

Note that the partial derivatives of $r$ with respect to Cartesian variables $x_i$, $i=1,2,3$ are given by

$$\frac{\partial r}{\partial x_i}=\frac{x_i}{r}$$

Multiplying by the corresponding Cartesian unit vector $\hat x_i$ and summing over $i$, we obtain

$$\sum_{i=1}^3 \hat x_i\frac{\partial r}{\partial x_i} =\frac{\vec r}{r}\tag 2$$

Using $(2)$ in $(1)$ yields

$$\nabla f(r)=f'(r)\nabla (r)$$

which was to be shown!