91Ó°ÊÓ

To find the partial derivative of \(p(x, y)\) with respect to \(x\), we will treat \(y\) as a constant. Thus, $$\frac{\partial p}{\partial x} = \frac{\partial}{\partial x}\left(\sqrt{12-4 x^{2}-y^{2}}\right)$$ To simplify finding the derivative, consider the transformation \(u = 12 - 4x^2 - y^2\). Then, we can rewrite the equation as \(p(x, y) = \sqrt{u}\). Using the chain rule: $$\frac{\partial p}{\partial x} = \frac{\partial p}{\partial u} \cdot \frac{\partial u}{\partial x}$$ Find the partial derivatives: $$\frac{\partial p}{\partial u} = \frac{1}{2 \sqrt{u}}$$ $$\frac{\partial u}{\partial x} = -8x$$ Now, we can find the partial derivative of \(p\) with respect to \(x\): $$\frac{\partial p}{\partial x} = \frac{1}{2 \sqrt{u}}(-8x) =\frac{-4x}{\sqrt{12 - 4x^2 - y^2}}$$

Similarly, to find the partial derivative of \(p(x, y)\) with respect to \(y\), we will treat \(x\) as a constant. We can reuse the transformation from Step 1 since \(u = 12 - 4x^2 - y^2\). Then using the chain rule again: $$\frac{\partial p}{\partial y} = \frac{\partial p}{\partial u} \cdot \frac{\partial u}{\partial y}$$ The partial derivative of \(u\) with respect to \(y\) is: $$\frac{\partial u}{\partial y} = -2y$$ Now, we can find the partial derivative of \(p\) with respect to \(y\): $$\frac{\partial p}{\partial y} = \frac{1}{2 \sqrt{u}}(-2y) =\frac{-y}{\sqrt{12 - 4x^2 - y^2}}$$

Now that we have found the partial derivatives of \(p(x, y)\) with respect to \(x\) and \(y\), we can write down the gradient: $$\nabla p(x, y) = \left(\frac{-4x}{\sqrt{12 - 4x^2 - y^2}}, \frac{-y}{\sqrt{12 - 4x^2 - y^2}}\right)$$

Lastly, we are asked to evaluate the gradient at point \(P(-1, -1)\): $$\nabla p(-1, -1) = \left(\frac{4}{\sqrt{12 - 4(-1)^2 - (-1)^2}}, \frac{1}{\sqrt{12 - 4(-1)^2 - (-1)^2}}\right) = \left(\frac{4}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)$$ Therefore, the gradient of the given function at point \(P(-1, -1)\) is \(\left(\frac{4}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right)\).