Is the partial of function with respect to another function, equal to the dot product of their gradients?

Question

I was working through a proof for a class I'm currently in, proving the identity of

$ r \frac{\partial f}{\partial r} = \vec x \cdot\nabla f $

where

$ \vec x \epsilon \mathbb{R}^n \\ r = |\vec x| $

And f is any real valued differentiable function.

One of the given identities to us by the TA when we asked about how to go about the proof was:

$ \frac{\partial f}{\partial r} = \nabla {f}\cdot \nabla {r} $

What is the name of the above partial to gradient identity mentioned by the TA? How would you prove it?

Answer 1

Recall that $\partial f/\partial r$ at a point $\vec x$ is the radial derivative, i.e., the directional derivative of $f$ in the direction of the unit vector $\vec x/\|\vec x\| = \vec x/r$. So $$\frac{\partial f}{\partial r} = \nabla f\cdot\frac{\vec x}r.$$ Multiply by $r$.

By the way, it makes no sense to take a partial derivative with respect to a function.

Answer 2

Might be worth noting it readily follows from the definition of the terms in spherical coordinates.

If $x=r\sin\phi\cos \theta, y=r\sin \phi \cos \theta $, and $z=r\cos \phi$

We also have position vector $\hat{x}=\vec{r}=r \hat{r}$

This part I wish I knew when I was first learning vector calculus, the relationship between the chain rule and the gradient.

The chain rule tells us $df = \frac{\partial f}{\partial r}dr + \frac{\partial f}{\partial \theta}d\theta + \frac{\partial f}{\partial \phi}d\phi$

By a coordinate free definition of gradients, we also have $df=\nabla f \cdot \vec{ds}$.

In spherical coordinates, $\vec{ds}= dr \hat{r} + r\sin \phi d\theta \hat{\theta} + r d\phi \hat{\phi}$.

Equating the two df's, we find that $\nabla f = \frac{\partial f}{\partial r}\hat{r} + \frac{1}{r\sin \phi}\frac{\partial f}{\partial \theta}\hat{\theta} +\frac{1}{r}\frac{\partial f}{\partial \phi} \hat{\phi}$

From this, it readily follows that $\hat{x} \cdot \nabla f=(r\hat{r}) \cdot (\frac{\partial f}{\partial r}\hat{r} + \frac{1}{r\sin \phi}\frac{\partial f}{\partial \theta}\hat{\theta} +\frac{1}{r}\frac{\partial f}{\partial \phi} \hat{\phi})=r\frac{\partial f}{\partial r}$

since $\hat{r} \cdot \hat{\theta}=\hat{r}\cdot \hat{\phi}=0$ and $\hat{r} \cdot \hat{r} =1 $.

$\nabla(r)=\hat{r}$ by definition, so $\nabla (r) \cdot \nabla f = \frac{\partial f}{\partial r}$

$df=\nabla f \cdot \vec{ds}$ is key to find the gradient in arbitrary coordinate systems. Extremely useful.