Calculating the Divergence of a Vector Field

[Karol]

Homework Statement

The question is to draw the function:
$$V=\frac {\boldsymbol{\hat {r}}}{r^2}$$
And to compute it's divergence: $\nabla \cdot V$

Homework Equations

$$\nabla\cdot V=\left ( \frac {\partial}{\partial x} \boldsymbol{\hat {x}}+\frac {\partial}{\partial y} \boldsymbol{\hat {y}}+\frac {\partial}{\partial z} \boldsymbol{\hat {z}}\right )\left ( v_x \boldsymbol{\hat {x}}+v_y \boldsymbol{\hat {y}}+v_z \boldsymbol{\hat {z}}\right )=\frac {\partial v_x}{\partial x}+\frac {\partial v_y}{\partial y}+\frac {\partial v_z}{\partial z}$$

The Attempt at a Solution

$$V=\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}+\frac {y}{x^2+y^2+z^2}\boldsymbol{\hat {y}}+\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}$$
The x component:
$$\frac {\partial}{\partial x}\left( \frac {x}{x^2+y^2+z^2}\right)=\frac {-2x^2+y^2+z^2}{(x^2+y^2+z^2)^2}$$
And the other components are similar
I don't think this is the answer, since i was told to expect something special, maybe a number?
The vector field is, to my opinion, as in the picture attached

 

[HallsofIvy]

Homework Statement

The question is to draw the function:
$$V=\frac {\boldsymbol{\hat {r}}}{r^2}$$
And to compute it's divergence: $\nabla \cdot V$

Homework Equations

$$\nabla\cdot V=\left ( \frac {\partial}{\partial x} \boldsymbol{\hat {x}}+\frac {\partial}{\partial y} \boldsymbol{\hat {y}}+\frac {\partial}{\partial z} \boldsymbol{\hat {z}}\right )\left ( v_x \boldsymbol{\hat {x}}+v_y \boldsymbol{\hat {y}}+v_z \boldsymbol{\hat {z}}\right )=\frac {\partial v_x}{\partial x}+\frac {\partial v_y}{\partial y}+\frac {\partial v_z}{\partial z}$$

The Attempt at a Solution

$$V=\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}+\frac {y}{x^2+y^2+z^2}\boldsymbol{\hat {y}}+\frac {x}{x^2+y^2+z^2}\boldsymbol{\hat {x}}$$
The x component:
$$\frac {\partial}{\partial x}\left( \frac {x}{x^2+y^2+z^2}\right)=\frac {-2x^2+y^2+z^2}{(x^2+y^2+z^2)^2}$$

This derivative is incorrect. And even if it were it wouldn't be "the answer" because you haven't calculated the other two partial derivatives.

I don't think this is the answer, since i was told to expect something special, maybe a number?
The vector field is, to my opinion, as in the picture attached

 

[Karol]

Why isn't the derivative correct?
The derivative of a fraction:
$$\frac {u}{v}=\frac {u'v-v'u}{v^2}$$
And, in our case:
$$\left ( \frac {x}{x^2+y^2+z^2} \right )'=\frac {1 \cdot (x^2+y^2+z^2)-2x \cdot x}{(x^2+y^2+z^2)^2}=\frac {-x^2+y^2+z^2}{(x^2+y^2+z^2)^2}$$

 

[HallsofIvy]

Yes, that is correct. And not what you had before.

Now calculate the other two partial derivatives and add them.