The force in an electrostatic field given by f(x, y, z) has the direction of the gradient. Find ∇f and its value at P.

The gradient of a function is defined as

$$\nabla=\frac{\partial}{\partial x}\textbf{i}+\frac{\partial}{\partial y}\textbf{j}= \Big(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\Big)$$

Let's find partial derivates.

$$\begin{align*} \frac{\partial}{\partial x} &= y\\ \frac{\partial}{\partial y} &=x \end{align*}$$

Consequently,

$$\boxed{ \nabla f=\Big(\frac{\partial}{\partial x},\frac{\partial}{\partial y}\Big)=(y, x)}$$

Since $p=(-4,5)$ we have:

$$\color{#c34632}{ \nabla f (p)= (5,-4)}$$

Given a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, it's gradient $\nabla f$ is a function from $\mathbb{R}^2$ to $\mathbb{R}^2$. At a point $(x,y)$, the direction of $\nabla f(x,y)$ points to the direction of the greatest growth of $f$, and it's magnitude tells us the speed of this growth.