Linking Gravitational Field Strength and Gravitational Potential confusion I want to ask a question about linking gravitational field strength g and gravitational potential V.
If I consider both quantities at a distance r from the centre of planet of mass M then I can see that they are closely related:
if $g = \frac{GM}{r^2}$ and $V = -\frac{GM}{r}$ then at a given point,
$$g = -\frac{-GM}{r}\times\frac{1}{r} = -\frac{V}{r}$$
so 
$$g = -\frac{V}{r}$$
Now, the book I am reading also states that I can link these variables through the graphs.
The rate of increase of the gravitational potential with distance r is known as the potential gradient, $\frac{\Delta V}{\Delta r}$ and that this value is equal to g, the gravitational field strength at that distance.

I see this is possible because using the equation we derived previously:
$$g = -\frac{V}{r}$$
$$-gr = V$$
$$-g = \frac{\Delta V}{\Delta R}$$
$$g = -\frac{\Delta V}{\Delta R}$$
but my question is this:
using the graph, why can I not just take a point on the line and use the formula $g = -\frac{V}{r}$ instead of differentiation using $g = -\frac{\Delta V}{\Delta R}$ to find the value of g at a distance r?
 A: In this case of gravitation force, you can indeed use $g=-\frac{V}{r}$ as well as differentiate V(r):
$$
\frac{dV(r)}{dr}=-GM\frac{d}{dr}(\frac{1}{r})=GM(\frac{1}{r^2})=g
$$
I guess the book is trying to establish the more general relationship between Force (F) and Energy (E). Broadly stating: 
$$
E=\int{}{}{F}dr ==> F=\frac{dE}{dr}
$$
A: Post the discussion in comments, here is my answer:
It is not correct to say that $g = -\frac{\partial V}{\partial r}$, where $g$ is the magnitude of the gravitational field. The full statement for a field derived from a potential is
$$
\boldsymbol g = -\nabla V
$$
The gravitational field of a point mass is
$$
\boldsymbol g = -\frac{GM}{r^2}\hat r
$$
for which
$$
\begin{gather}
V = -\frac{GM}{r} \\
-\nabla V = -\frac{\partial V}{\partial r}\hat r = \frac{V}{r}\hat r
\end{gather}
$$
Taking magnitudes, we see that 
$$
\begin{gather}
g = \frac{\partial V}{\partial r} = -\frac{V}{r} \\
V = -gr \\
\end{gather}
$$
Differentiating,
$$
\frac{\partial V}{\partial r} = -\frac{\partial (gr)}{\partial r} = -g - r\frac{\partial g}{\partial r} = -g + 2g = g
$$
which is consistent.
It is important to note that 
$$
\frac{\partial V}{\partial r} = -\frac{V}{r}
$$
is true only if
$$
V = \frac{k}{r}
$$
for some constant $k$.
