Why no negative sign in gradient for Newton's method

[zmalone]

I'm watching a lecture on Newton's method with n-dimensions but I am kind of hung up on why the professor did not use the negative sign while taking the first gradient? Is there a rule that explains this or something that I'm forgetting? The rest makes sense but highlighted in red is the part I am confused on if anyone can clear that up I'd appreciate it, thanks!

Where g(x,y) = 1-(x-1)^4-(y-1)^4

local maximum at (1,1) ; critical point at (1,1)

Gradient of g(x,y):

F(x,y,) = [Dg(x,y,)]transpose = [4(x-1)^3 4(y-1)^3]transpose
Why not [-4(x-1)^3 -4(y-1)^3]?

Gradient of F(x,y):

DF(x,y) =
12(x-1)^2 0
0 12(y-1)^2

Screen shot which is probably easier to read:

 

[vanhees71]

The gradient in Cartesian components is, of course, given by
$$\vec{\nabla} g=(\partial_x g,\partial_y g)=(-4(x-1)^3,-4 (y-1)^3)$$
and the Hesse Matrix by
$$H_{ij}=\partial_i \partial_j g=\mathrm{diag}(-2(x-1)^2,-12(y-1)^2).$$