Showing a vector field is imcompressible

[question dude]

Homework Statement

Homework Equations

The Attempt at a Solution

As you can see, the solution is shown just below the question.

Essentially, I don't understand how the x, y and z component of the vector field has been separated because the numerator of the vector field's fraction is: (x^2 + y^2 + z^2)^(1/2)

It seems like they've just taken x^2 out and square rooted it to get 'x', but you can't do that, can you?

 

[DEvens]

https://en.wikipedia.org/wiki/Product_rule
https://en.wikipedia.org/wiki/Chain_rule

 

[LCKurtz]

Homework Statement

Homework Equations

The Attempt at a Solution

As you can see, the solution is shown just below the question.

Essentially, I don't understand how the x, y and z component of the vector field has been separated because the numerator of the vector field's fraction is: (x^2 + y^2 + z^2)^(1/2)

It seems like they've just taken x^2 out and square rooted it to get 'x', but you can't do that, can you?

That isn't what they have done. The numerator is bold faced $\bf{r}$ and the denominator is $r$. The first is the vector and the second its magnitude. r = xi + yj + zk.

 

[question dude]

That isn't what they have done. The numerator is bold faced $\bf{r}$ and the denominator is $r$. The first is the vector and the second its magnitude. r = xi + yj + zk.

Do I treat the non-bold r as a constant?

If sub in bold r, I get:

G = [(x^2 + y^2 + z^2)^0.5 ] / 4*pi*r^3

 

[LCKurtz]

Do I treat the non-bold r as a constant?

If sub in bold r, I get:

G = [(x^2 + y^2 + z^2)^0.5 ] / 4*pi*r^3

I just noticed in your graphic under (b) they have ${\bf r} = \sqrt{x^2+y^2+z^2}$ That should not have been a bold face r. The bold face r represents the vector and the plain r its magnitude, which is not constant.