We know that

Gradients traditionally come in the form of G(x,y) = A(x,y)

**i**+B(x,y)

**j** -- yet, not every function in this form is a Gradient. This becomes critical if one is trying to

reconstruct a function from what's believed to be the Gradient.
We know that in the above form, A(x,y) is the

partial in respect to x (

*d*f/

*d*x) and B(x,y) is the partial in respect to y (

*d*f/

*d*y) and for any function continuously

differentiable

*d*f/*d*x*d*y = *d*f/*d*y*d*x

Hence, if we differentiate A(x,y)(*d*f/*d*x) in respect to y we attain *d*f/*d*x*d*y; if we differentiate B(x,y)(*d*f/*d*y) in respect to y we get *d*f/*d*y*d*x. Setting these equal, we are able to tell if the function is indeed a gradient.

Drawing an example from a previous node:

g(x,y)=(4x^{3}y^{3} - 3x^{2})**i** + (3x^{4}y^{2} + cos 2y)**j**

*d*f/*d*x(x,y) = (4x^{3}y^{3} - 3x^{2})

*d*f/*d*y(x,y) = (3x^{4}y^{2} + cos 2y)

*d*f/*d*x*d*y = 12x^{3}y^{2}

*d*f/*d*y*d*x = 12x^{3}y^{2}

12x^{3}y^{2} = 12x^{3}y^{2}

So the function is a Gradient.

*ariels says* OK then. But a function

can be a gradient without being further

differentiable, and still possess partial

derivatives.

Very true. In this case... I don't exactly know what you would do, except trying to reconstruct the function and seeing if anything of value comes out.