Calculating Gradient Vector at Point S: x=4, y=8, z=-6

[Air]

Homework Statement

Calculate the gradient vector at the point $$S$$ for the function, $$f(x,y,z)=x-\sqrt{z^2 - y^2}; S(x,y,z)=(4, 8, -6)$$.


2. The attempt at a solution
$$\frac{\partial f}{\partial x} = 1$$
$$\frac{\partial f}{\partial y} = \frac{y}{\sqrt{z^2-y^2}}$$
$$\frac{\partial f}{\partial z} = -\frac{z}{\sqrt{z^2-y^2}}$$
Therefore at Point $$S$$:
$$\frac{\partial f}{\partial x} = 1$$
$$\frac{\partial f}{\partial y} = \frac{-6}{\sqrt{36-64}} = \frac{-6}{\sqrt{-28}}$$


3. The problem that I'm having
When I put the point values of Point $$S$$ into the equation, I get a negative square root in the denominator. Where am I going wrong? Thanks in advance.

 

[Matterwave]

Indeed, this is weird. Have you double checked you copied the problem down right?

We can see that the scalar field becomes imaginary at that point as well (you will also get a negative sign under the square root).