evalf and subs in sympy on single variable expression returns expression instead of expected float value

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I'm new to sympy and I'm trying to use it to get the values of higher order Greeks of options (basically higher order derivatives). My goal is to do a Taylor series expansion. The function in question is the first derivative.

f(x) = N(d1)

N(d1) is the P(X <= d1) of a standard normal distribution. d1 in turn is another function of x (x in this case is the price of the stock to anybody who's interested).

d1 = (np.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*np.sqrt(0.5))

As you can see, d1 is a function of only x. This is what I have tried so far.

import sympy as sp
from math import pi
from sympy.stats import Normal,P

x = sp.symbols('x')
u = (sp.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*np.sqrt(0.5))
N = Normal('N',0,1)
f = sp.simplify(P(N <= u))
print(f.evalf(subs={x:100})) # This should be 0.5155
f1 = sp.simplify(sp.diff(f,x))
f1.evalf(subs={x:100}) # This should also return a float value

The last line of code however returns an expression, not a float value as I expected like in the case with f. I feel like I'm making a very simple mistake but I can't find out why. I'd appreciate any help.

Thanks.

sympy python
1 Answers

If you define x with positive=True (which is implied by the log in the definition of u assuming u is real which is implied by the definition of f) it looks like you get almost the expected result (also using f1.subs({x:100}) in the version without the positive x assumption shows the trouble is with unevaluated polar_lift(0) terms):

import sympy as sp
from sympy.stats import Normal, P

x = sp.symbols('x', positive=True)
u = (sp.log(x/100) + (0.01 + 0.5*0.11**2)*0.5)/(0.11*sp.sqrt(0.5)) # changed np to sp
N = Normal('N',0,1)
f = sp.simplify(P(N <= u))
print(f.evalf(subs={x:100})) # 0.541087287864516
f1 = sp.simplify(sp.diff(f,x))
print(f1.evalf(subs={x:100})) # 0.0510177033783834
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