How do you compute the gradient of the function `p(x,y)=sqrt(24-4x^2-y^2)` and then evaluate it at the point (-2,1)?

This is conceptually very simple, so the only "difficult" part will be the calculations: if `n` is the number of variabiles of a function (in your case thus `n=2`, then the gradient of the function is a vector of length `n`, whose elements are the derivatives with respect to each variables, so if you have `f(x_1,...,x_n)`, the gradient will be the vector
`({\partial f}/{\partial x_1},..,{\partial f}/{\partial x_n})`

So, the gradient of your function `p` will be the 2-dimensional vector
`({\partial p}/{\partial x},{\partial p}/{\partial y})`.

Let's calculate the two derivatives: by the chain rule, if we need to derive an expression such `\sqrt{f(x,y)}`, the derivative will be `{f'(x,y)}/{2\sqrt{f(x,y)}}`, where of course `f'(x,y)` is the derivative with respect to the right variable. So, deriving with respect to `x` (which means that we must consider `y` as a constant) , we get `f'(x,y)=-8x`.

Deriving with respect to `y`, we get instead `-2y`. So, the gradient vector will be

`( {-8x}/{2\sqrt{24-4x^2-y^2}}, {-2y}/{2\sqrt{24-4x^2-y^2}})`

We can simplify both terms, obtaining

`( {-4x}/{\sqrt{24-4x^2-y^2}}, {-y}/{\sqrt{24-4x^2-y^2}})`

To evaluate this vector, we only need to plug in the values, if `x=-2` and `y=1`, we get

`(8/sqrt(7), -1/sqrt(7) )`