Determine the gradient vector of the function f(x,y)=x^3*y^2-2x at the point (2,4)?

Since the function is of 2
variables, the gradient of the function is the vector
function:

Grad f(x,y) = [df(x,y)/dx]*i +
[df(x,y)/dy]*j

[df(x,y)/dx] and [df(x,y)/dy] are the
partial derivatives of the function.

To determine
df(x,y)/dx, we'll differentiate the function with respect to x, assuming that y is a
constant.

df(x,y)/dx =3x^2*y^2 -
2

To determine df(x,y)/dy, we'll differentiate the function
with respect to y, assuming that x is a
constant.

df(x,y)/dy =
2y*x^3

Grad f(x,y) = (3x^2*y^2 - 2)*i +
(2y*x^3)*j

Grad f(2,4) =(3*2^2*4^2 - 2)*i +
(2*4*2^3)*j

Grad f(2,4) = 190*i +
64*j

The gradient vector of the given
function, at the point (2,4), is:Grad f(2,4) = 190*i +
64*j.