For a position vectorthe norm of the vector can be defined as . Given a function ,  its gradient is

The correct answer is:

$\frac{\overrightarrow{r}}{\overrightarrow{r}·\overrightarrow{r}}$

Step-by-step Derivation:

1. We are given the position vector:
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$
and its magnitude (norm):
$r=\left|\overrightarrow{r}\right|=\sqrt{x^2+y^2+z^2}$

Squaring both sides gives:
$r^2=x^2+y^2+z^2=\overrightarrow{r}·\overrightarrow{r}$

2. We compute the partial derivatives of $r$ with respect to $x$, $y$, and $z$ using implicit differentiation:
$2r\frac{\partial r}{\partial x}=2x\Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}$
By symmetry, the partial derivatives with respect to $y$ and $z$ are:
$\frac{\partial r}{\partial y}=\frac{y}{r}$
and
$\frac{\partial r}{\partial z}=\frac{z}{r}$

3. Next, we find the gradient of the function $\varphi =lnr$. The gradient operator in Cartesian coordinates is defined as:
$\nabla \varphi =\frac{\partial \varphi }{\partial x}\hat{i}+\frac{\partial \varphi }{\partial y}\hat{j}+\frac{\partial \varphi }{\partial z}\hat{k}$

Using the chain rule, we compute each component:
$\frac{\partial \varphi }{\partial x}=\frac{\partial }{\partial x}(lnr)=\frac{1}{r}\frac{\partial r}{\partial x}=\frac{1}{r}·\frac{x}{r}=\frac{x}{r^2}$
Similarly for $y$ and $z$:
$\frac{\partial \varphi }{\partial y}=\frac{y}{r^2}$
and
$\frac{\partial \varphi }{\partial z}=\frac{z}{r^2}$

4. Substitute these components back into the gradient formula:
$\nabla \varphi =\frac{x}{r^2}\hat{i}+\frac{y}{r^2}\hat{j}+\frac{z}{r^2}\hat{k}$
Factor out the common term $\frac{1}{r^2}$:
$\nabla \varphi =\frac{1}{r^2}(x\hat{i}+y\hat{j}+z\hat{k})$

5. Since $x\hat{i}+y\hat{j}+z\hat{k}=\overrightarrow{r}$ and $r^2=\overrightarrow{r}·\overrightarrow{r}$, we obtain:
$\nabla \varphi =\frac{\overrightarrow{r}}{\overrightarrow{r}·\overrightarrow{r}}$