What would the plot of partial derivative look like?

Question

Suppose, I have the following function:

$$f(x,y) = -x^2 - y^2 + 4 \qquad (1)$$

Partial derivative of $(1)$ w.t.r. $x$ would be:

$$\frac{\partial f}{\partial x} = -2x \qquad (2)$$

Partial derivative of $(1)$ w.t.r. $y$ would be:

$$\frac{\partial f}{\partial y} = -2y \qquad (3)$$

So, the partial derivative of $(1)$ w.r.t. both $x$ and $y$ would be:

$$ \text{Gradient of } f = \nabla f = \begin{bmatrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\end{bmatrix} = \begin{bmatrix}-2x \\ -2y\end{bmatrix} \qquad (4)$$

I have two questions here:

1. Is the above description correct?
2. What would the graph/plot of $(4)$ look like?

Answer 1

The important result you should remember here is

If the function $f$ is differentiable, the gradient of $f$ at a point is either zero, or perpendicular to the level set of $f$ at that point.

The level sets of $f$ are circles. For example $0 = -x^2 -y^2 +4 $ is a circle of radius $2$. And the gradient must be perpendicular to that circle, so it is a radius. Here's a sketch

Arrows represent the gradient, and colors the function. The circles are level sets