# Another vector identity question (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### Benny

Hi, I'm stuck another vector identity question. It's of a different kind to the other one I asked about and looks so much easier but I just can't see what I need to do.

I am told to use standard identities to deduce the following result. The standard identities being referred to are listed in my notes and as I'm sure you can appreciate, it would be a PITA to type them all out. It's just the usual list comprised of various combinations of grad, div curl, and higher dimensional equivalents of product and quotient rules.

Anyway here is the identity I need to show.

$$\nabla \left( {r^n } \right) = nr^{n - 2} \mathop r\limits^ \to$$

where $r = \left\| {\mathop r\limits^ \to } \right\|$ and r = xi + yj + zk.

I haven't written this down but from a quick look at the relation, I would've thought that the RHS should have an (n-1) rather than an (n-2) as given. In any case, grad(r^n) looks so simple to compute that I don't see how I do it without just writing out an explicit expression for the gradient. I can't think of any identities which could help - I'm told to use identities for this question.

Any help would be good thanks.

Edit: The (n-2) is probably right now that I think about it. Anyway I'm off to sleep.

Last edited:

#### Galileo

Homework Helper
If you write it with the unit vector $\hat r$ it's:

$$\nabla (r^n)=nr^{n-1}\hat r$$

which ofcourse reminds you of $d/dx (x^n)=nx^{n-1}$, so this is, as you can guess, proved in a very similar way.

#### Benny

Thanks for the suggestion, I'll see what I can come up with.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving