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

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The gradient vector is a vector whose first component is the derivative of f(x, y) with respect to x and the second is the derivative of f(x, y) with respect to y.

Here f(x, y) = x^3*y^2 - 2x

derivative with respect to x is 3x^2*y^2 - 2

derivative with respect to y is 2y*x^3

At the point (2, 4)

3x^2*y^2 - 2 = 3*2^2*4^2 - 2 = 190

2y*x^3 = 64

**The required vector isĀ < 190 , 64>**

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.**