What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$?

Question

What is the gradient of $f(x, y, z) = \sqrt{x^2+y^2+z^2}$?

I know that $\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle$ or equivalently $\nabla f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$.

Eventually, I am going to evaluate this gradient at a point and then dot it with some given unit vector to find a directional derivative. I think maybe what is troubling me is the radical in $f$. Will I have to use the chain rule in order to find the three partial derivatives?

I tried this:

$$\frac{\partial f}{\partial x} = \frac{1}{2}(x^2 + y^2 + z^2)^{-\frac{1}{2}}2x$$ $$= \frac{x}{\sqrt{x^2 + y^2 + z^2}}$$

That result looks ugly to me. However, if I did that partial derivative correctly, then I think I know what I am doing.

Answer 1

$$\nabla \|x\| = \frac{x}{\|x\|} $$ where $x = (x_1,x_2,\ldots, x_n)$ and $\|x\| = \sqrt{x_1^2+x_2^2+\ldots + x_n^2}$.

Answer 2

I'm fond of deriving this result as such. Note that $f(\mathbf{r})^2=x^2+y^2+z^2=\mathbf{r}\cdot \mathbf{r}$ by the definition of the dot product. Then the product rule for gradients implies $\nabla (f(\mathbf{r})^2)=2f(r)\nabla f(\mathbf{r})$ and therefore $$\nabla f(\mathbf{r})=\dfrac{\nabla (f(\mathbf{r})^2)}{2f(\mathbf{r})}=\dfrac{\nabla(x^2+y^2+z^2)}{2f(\mathbf{r})}=\dfrac{\langle 2x,2y,2z\rangle}{2f(\mathbf{r})}=\dfrac{\langle x,y,z\rangle}{\sqrt{x^2+y^2+z^2}}$$

In this form, the seemingly 'ugly' components become symmetric. Moreover, note that 1) this is a unit vector, 2) the denominator is just $\mathbf{r}$. So $\nabla f(\mathbf{r})=\dfrac{\mathbf{r}}{\sqrt{x^2+y^2+z^2}}=\dfrac{\mathbf{r}}{|\mathbf{r}|}=\hat{\mathbf{r}},$ i.e. the gradient is always a unit vector pointing outwards from the origin. That seems rather nice to me!

Answer 3

So you're asking about what is $\nabla_{\vec{v}}f$ where $f = \sqrt{x^2 + y^2 + z^2}$. First, lets organize the answer: $\nabla_{\vec{v}}f = \begin{aligned} \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \end{bmatrix} \end{aligned}$ $\cdot \, \vec{v}$
Now, lets solve for $\frac{\partial f}{ \partial x}$.

$$\frac{\partial}{\partial x}\left(\sqrt{x^2 + \overbrace{y^2 + z^2}^{\mathrm{a\ constant}}}\right)$$ Now, lets start, also just for the sake of space, I'll make this substitution $y^2 + z^2 =c$, so we have $$ \frac{\partial f }{\partial x} = \frac{d}{dx}\sqrt{x^2 + c} $$

so $g(x) = \sqrt{x},\, h(x) = x^2 + c$

I am going to skip some steps. $$g'(x) = \frac{1}{2\sqrt{x}}, \qquad h'(x) = 2x$$

$$\frac{d}{dx} = g'(h(x)) \bullet h'(x) \rightarrow \frac{4x}{\sqrt{x^2 \, + \, c }} \rightarrow \mathbf{\frac{x}{\sqrt{x^2 + y^2 + z^2}}}$$

Then, $$\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \, \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \, \frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}.$$

Okay, so now let's find a directional derivative! But first, let's go to the gradient. say $\vec{v} = \begin{aligned} \begin{bmatrix} a \\ b \\ c \end{bmatrix} \end{aligned}$

$$\begin{aligned} \begin{bmatrix} \frac{x}{\sqrt{x^2 + y^2 + z^2}} \\ \frac{y}{\sqrt{x^2 + y^2 + z^2}} \\ \frac{z}{\sqrt{x^2 + y^2 + z^2}} \end{bmatrix} \cdot \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} \end{aligned} = \frac{xa \,+ \, yb \, + \, zc}{\sqrt{x^2 \, +\, y^2 \, + \, z^2}}$$

I hope I helped you out on this page.