Understanding what a gradient vector is Definition: let $A$ be an open set in a Euclidean space $X$. Let $f : A \to \mathbb R$ be a real valued differentiable function. Then,for each $a \in A$, there is a unique vector $\nabla f(a) x \in X$ such that $f'(a) x = \langle \nabla f(a), x \rangle$ for all $x \in X$. The vector $\nabla f(a) \in X$ is called the gradient vector of $f$ at $a\in ..$
I still can't understand what the gradient is? For what purpose is it used?
Please if anyone can provide some intuition on it.
 A: The gradient points to the direction of faster growing of the function. See The Gradient and Directional Derivative.
A: Imagine a landscape, where $x$ is west-east, and $y$ is north-south. Let $f(x,y)$ be the height at that point. If we take the gradient of $f$, we get a function from $\mathbb{R}^2$ to $\mathbb{R}^2$. At $(x,y)$, $\nabla f(x,y)$ points "uphill"; that is, it points in the direction of steepest ascent.
This is useful for physics. If you have some kind of potential energy function, taking the gradient will get you the corresponding force field. For example, consider a point mass $M$ fixed at the origin, and a smaller mass $m$ elsewhere. The potential energy of the small particle is $U = -G \frac{Mm}{r}$. If you take the gradient, you'll get $\vec{F} = G \frac{Mm}{r^2}$, which corresponds* with the results from physics.
We have to convert to rectangular coordinates, take the gradient, then convert back to polar.
$$
\begin{align*}
U &= -G \frac{Mm}{r} \\
&= -G \frac{Mm}{\sqrt{x^2 + y^2 + z^2}} \\
\frac{\partial U}{\partial x} &= G \frac{Mm}{2 (x^2 + y^2 + z^2)^{3/2}} \cdot 2x = G \frac{Mm \cdot x}{(x^2 + y^2 + z^2)^{3/2}} \\
\frac{\partial U}{\partial y} &= G \frac{Mm \cdot y}{(x^2 + y^2 + z^2)^{3/2}} \\
\frac{\partial U}{\partial z} &= G \frac{Mm \cdot z}{(x^2 + y^2 + z^2)^{3/2}} \\
\nabla U &= \left\langle \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right\rangle \\
&= G \frac{Mm}{(x^2 + y^2 + z^2)^{3/2}} \langle x, y, z \rangle \\
&= G \frac{Mm}{x^2 + y^2 + z^2} \frac{\langle x, y, z \rangle}{\sqrt{x^2 + y^2 + z^2}} \\
&= G \frac{Mm}{r^2} \hat{r}
\end{align*}
$$

*Really, we should take the negative of the gradient, because in physics, we want our force vectors pointing downhill, towards lower energy. But flipping the sign is easy.
