We know that Gradients
traditionally come in the form of G(x,y) = A(x,y)i
-- 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
x) and B(x,y) is the partial in respect to y (d
y) 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
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.