Height differences from relative uncertainty of gravimeters

[AdrianMachin]

Homework Statement

The best relative gravimeters have a relative uncertainty of 10^-12, that corresponds to a height difference of 3 µm.

Homework Equations

g∝(1/r²)
The local gravitational acceleration g outside the Earth is proportional to 1/r², which means (Δg)/g = -2 (Δr)/r. With (Δg)/g = 10^-12 we get Δr = 0.5 * 10^-12 r where r is the radius of Earth.

The Attempt at a Solution

I've tried to get to the equation I highlighted in red, but I fail. I know it has derived from taking derivatives, but my derivatives result in different equations.
g∝1/r² → g=k(1/r²) → ?

 

[BvU]

The rule is derived from $f(x+\Delta x) = f(x) + f'(x)\Delta x\ + {\mathcal O} (\Delta x)^2$ so you get $$
\Delta g = g(x+\Delta x ) - g(x) \approx g'(x) \Delta x \Rightarrow \\ = {-2\over x^3} \Delta x \Rightarrow {\Delta g\over g} = -2 {\Delta x\over x}$$

We usually report the standard deviation which is positive.

 

[AdrianMachin]

The rule is derived from $f(x+\Delta x) = f(x) + f'(x)\Delta x\ + {\mathcal O} (\Delta x)^2$ so you get $$
\Delta g = g(x+\Delta x ) - g(x) \approx g'(x) \Delta x \Rightarrow \\ = {-2\over x^3} \Delta x \Rightarrow {\Delta g\over g} = -2 {\Delta x\over x}$$

We usually report the standard deviation which is positive.

Thanks. What's the name of that rule and the equation you wrote?
I've totally forgotten it, so I'm asking for a reference to learn and review.

 

[BvU]

differentation ...