Quick question to clarify my understanding of how to find the gradient of a function: In practice exam 2b, Problem #1: (x^2)(y^2)-x

Question

anonymous · Answer

all right, to find the gradient of this function, $$f(x,y) = x^2y^2-x$$

anonymous · Answer

you find the partial derivative of f with regard to x, and then with regard to y

anonymous · Answer

$$<\partial f/\partial x, \partial f/\partial y>$$

anonymous · Answer

all right, so first, partial w/re to x is 2x*y^2-1

anonymous · Answer

and partial w/re to y is 2x^2*y

anonymous · Answer

looks good, so format would be: $$<2xy^2-1,  2x^2y>$$

anonymous · Answer

right, excellent, just wanted to make sure my brain was working properly ;)

anonymous · Answer

now you can just plug in the point to consider like in the prac. exam problem

anonymous · Answer

idk.