Finding $\nabla r^n$

Question

Find $\nabla r^n$ where $r= \sqrt{x^2+y^2+z^2}$.

So $r^n = (x^2+y^2+z^2)^{n/2}$. Then $$\frac{\partial r^n}{\partial x} = 2x\cdot\frac{n}{2}\cdot(x^2+y^2+z^2)^{n/2 - 1} = nx(x^2+y^2+z^2)^{(n-2)/2} = nxr^{n-2},$$

similarly $$\frac{\partial r^n}{\partial y} = nyr^{n-2}, \quad \frac{\partial r^n}{\partial z} = nzr^{n-2}.$$

Thus $$\nabla r^n = nr^{n-2} (x\mathbf{i} + y\mathbf{j} + z\mathbf{k}) = nr^{n-2}\mathbf{r}.$$

However the correct answer is $$\nabla r^n=nr^{n-1} \hat{\mathbf{r}}.$$

I don't really understand where I have gone wrong.

Are you sure the correct answer is $nr^{n-1}\mathbf r$ or $nr^{n-1}\mathbf{\hat r}$? — kennytm, May 08 '15 at 16:21
So since $\mathbf r = r \hat {\mathbf r}$, you're answer is OK, no? — Robert Lewis, May 08 '15 at 16:28
I edited your question in accord with these comments. Hope it's OK. Cheers! — Robert Lewis, May 08 '15 at 17:28

score 3 · Accepted Answer · answered May 08 '15 at 18:42

The OP user2850514's solution

$\nabla r^n = nr^{n - 2} \mathbf r \tag{1}$

is in fact correct since

$\mathbf r = r \hat{\mathbf r}; \tag{2}$

it is simply written in a slightly different fashion.

I do this one like this:

For differentiable $g: \Omega \to \Bbb R$ and $f:I \to \Bbb R$ with $\Omega$ open in $\Bbb R^m$, $I$ open in $\Bbb R$ and $g(\Omega) \subset I$, we have for $x \in \Omega$,

$\nabla (f(g))(x) = \dfrac{df(g(x))}{dg}(g(x)) \nabla g(x); \tag{3}$

this identity is well-known and is really just the chain rule, as may be seen by looking at it in coordinates $x = (x_1, x_2, \ldots, x_m)$ in $\Omega$:

$(\nabla(f(g))(x))_k = \dfrac{\partial(f(g)(x))}{\partial x_k} = \dfrac{d(f(g))}{dg}(g(x)) \dfrac{\partial g(x)}{\partial x_k} = \dfrac{d(f(g))}{dg} (g(x))(\nabla g(x))_k; \tag{4}$

(4) is precisely (3), coordinate-by-coordinate; thus (4) $\Rightarrow$ (3). Also,

$\nabla r = \hat{\mathbf r}; \tag{5}$

again we use the coordinates:

$(\nabla r)_k = \dfrac{\partial r}{\partial x_k} = \dfrac{\partial \sqrt{\sum_1^m x_i^2}}{\partial x_k} =\dfrac{1}{2}(\sqrt{\sum_1^m x_i^2})^{-1/2}(2x_k) = \dfrac{x_k}{r};\tag{6}$

but $x_k / r$ is just the $k$-th component of $\hat{\mathbf r}$, the unit vector field pointing in the $\mathbf r$ direction; hence (5) binds.

Now taking $f(r) = r^n$ and $g(x) = r$ we have

$\nabla r^n = nr^{n - 1} \nabla r = nr^{n - 1} \hat{\mathbf r}; \tag{7}$

that does it!

Of course, in solving such problems I really go directly from (3), (5) to (7); (3) and (5) are standard, useful identities living in my head (that is, memory); I don't re-derive them over and over. But I thought the details might help flesh things out here.

Finding $\nabla r^n$

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