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I am working with some code that generates an equation with a number of derivatives. I do not know the order of the derivatives or the order. If I was working with something algebraic, for example:

a, b, c = var('a,b,c')
eqn = a**2 + 3*b + c

I could use something like

coeffs = eqn.coefficients(a) highest_order = coeffs[-1][1]

to find out the highest order of a. It would also be okay if i could figure out the order of every derivative of a given function in an equation. I can do this if I know before-hand what the order of the derivative is function('f') eqn += f(a,b,c).diff(a,b) eqn.find(f(a,b,c).diff(a,b)) and see that in fact that derivative is there somewhere. Anyone have any ideas?

I am working with some code that generates an equation with a number of derivatives. I do not know the order of the derivatives or the order. If I was working with something algebraic, for example:

a, b, c = var('a,b,c')
eqn = a**2 + 3*b + c

I could use something like

coeffs = eqn.coefficients(a)
 highest_order = coeffs[-1][1]
coeffs[-1][1]

function('f')
 eqn += f(a,b,c).diff(a,b)
 eqn.find(f(a,b,c).diff(a,b))

I am working with some code that generates an equation with a number of derivatives. I do not know the order of the derivatives or the order. If I was working with something algebraic, for example:

a, b, c = var('a,b,c')
eqn = a**2 + 3*b + c

I could use something like

coeffs = eqn.coefficients(a)
highest_order = coeffs[-1][1]

to find out the highest order of a. It would also be okay if i could figure out the order of every derivative of a given function in an equation. I can do this if I know before-hand what the order of the derivative is

function('f')
eqn += f(a,b,c).diff(a,b)
eqn.find(f(a,b,c).diff(a,b))

and see that in fact that derivative is there somewhere. Anyone But what I really want, is for

eqn = f(a,b,c).diff(a,b) + b*f(a,b,c).diff(b,3) + c

to have any ideas?something that (a) tells me 'eqn has [D[0,1](f)(a,b,c), D[1,1,1](f)(a,b,c)]' or, (b) 'the highest order derivative of eqn is D[1,1,1](f)(a,b,c)'.