Recognizing a Gradient

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 (df/dx) and B(x,y) is the partial in respect to y (df/dy) and for any function continuously differentiable

df/dxdy = df/dydx

Hence, if we differentiate A(x,y)(df/dx) in respect to y we attain df/dxdy; if we differentiate B(x,y)(df/dy) in respect to y we get df/dydx. 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)=(4x³y³ - 3x²)i + (3x⁴y² + cos 2y)j

df/dx(x,y) = (4x³y³ - 3x²)
df/dy(x,y) = (3x⁴y² + cos 2y)

df/dxdy = 12x³y²
df/dydx = 12x³y²

12x³y² = 12x³y²

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.