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?

  • 1
    $\begingroup$ The textbook uses the chain rule: $df/dx=(df/dr)\cdot (dr/dx)$. $\endgroup$
    – MonadBoy
    Mar 11, 2016 at 16:30
  • $\begingroup$ @A.Sh The dot is the dot product other normal multiplication? $\endgroup$ Mar 11, 2016 at 16:47
  • $\begingroup$ I meant ordinary multiplication, i.e. "scalar times scalar". As in the answers posted, just apply it to each variable (x,y,z) to obtain each respective entry of the gradient vector. Sorry for any eventual confusion. $\endgroup$
    – MonadBoy
    Mar 11, 2016 at 19:25

3 Answers 3


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.

  • $\begingroup$ Thank you really much. You just saved me after an hour long searching. $\endgroup$ Mar 11, 2016 at 16:52

It is a consequence of the derivative of function.

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

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


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!


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