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

The correct option is:
$$\nabla \varphi =\frac{\overrightarrow{r}}{\overrightarrow{r}·\overrightarrow{r}}$$

Step-by-Step Derivation:
1. We are given the position vector from the image:
$$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
The norm (or magnitude) of this vector is:
$$\left|\overrightarrow{r}\right|=\sqrt{x^2+y^2+z^2}$$

2. We are given the function:
$$\varphi =ln\left|\overrightarrow{r}\right|=ln\left(\sqrt{x^2+y^2+z^2}\right)$$
Using logarithmic properties, we can rewrite the square root power of 1/2 as a coefficient:
$$\varphi =\frac{1}{2}$$

3. The gradient of the scalar function $\varphi$ is defined as:
$$\nabla \varphi =\frac{\partial \varphi }{\partial x}\hat{i}+\frac{\partial \varphi }{\partial y}\hat{j}+\frac{\partial \varphi }{\partial z}\hat{k}$$

4. Calculate the partial derivative with respect to $x$ using the chain rule:
$$\frac{\partial \varphi }{\partial x}=\frac{1}{2}$$
$$\frac{\partial \varphi }{\partial x}=\frac{1}{2}$$
By symmetry, the partial derivatives with respect to $y$ and $z$ are:
$$\frac{\partial \varphi }{\partial y}=\frac{y}{x^2+y^2+z^2}$$
$$\frac{\partial \varphi }{\partial z}=\frac{z}{x^2+y^2+z^2}$$

5. Substitute the partial derivatives back into the gradient formula:
$$\nabla \varphi =\left(\frac{x}{x^2+y^2+z^2}\right)\hat{i}+\left(\frac{y}{x^2+y^2+z^2}\right)\hat{j}+\left(\frac{z}{x^2+y^2+z^2}\right)\hat{k}$$
Combine the terms under a single common denominator:
$$\nabla \varphi =\frac{x\hat{i}+y\hat{j}+z\hat{k}}{x^2+y^2+z^2}$$

6. Simplify using vector notations:
The numerator is the position vector $\overrightarrow{r}$ itself:
$$x\hat{i}+y\hat{j}+z\hat{k}=\overrightarrow{r}$$
The denominator is the square of the magnitude of $\overrightarrow{r}$, which is equivalent to the dot product of $\overrightarrow{r}$ with itself:
$$x^2+y^2+z^2={\left|\overrightarrow{r}\right|}^2=\overrightarrow{r}·\overrightarrow{r}$$
Substituting these expressions gives the final result:
$$\nabla \varphi =\frac{\overrightarrow{r}}{\overrightarrow{r}·\overrightarrow{r}}$$