SymPy 在区分隐式 function 时没有简化“足够”

Question

I am trying to find the 1st and 2nd derivatives of the implicit function y = f(x)
which is defined by the equation: exp(sin(x)) - x * exp(sin(y)) = 0

SymPy calculates the 1st derivative and gives this answer:

But this expression can be written much simpler as:

(x * cos(x) - 1) / (x * cos(y))

using the fact that x = exp(sin(x)-sin(y))

The answer given for the 2nd derivative is also quite complicated.

Of course the second derivative can also be simplified
quite a lot using the same fact x = exp(sin(x)-sin(y)) .

How can I make/force SymPy apply these additional simplifications?
Is that possible even?

Here is my script.

#!/usr/bin/env python
# coding: utf-8

# ### Differentiating an implicit function using SymPy

# In[1]:


import sympy as sp


# In[2]:


sp.__version__


# In[3]:


sp.init_printing(use_latex='mathjax')  # use pretty mathjax output


# In[4]:


sp.var('x y z')

F = sp.exp(sp.sin(x)) - x * sp.exp(sp.sin(y))


# In[5]:


f1 = sp.idiff( F, y, x ) # First derivative of y w.r.t. x
f1


# In[6]:


sp.simplify(f1)


# In[7]:


f2 = sp.idiff( F, y, x, 2) # Second derivative of y w.r.t. x
f2

sp.simplify(f2)


# In[ ]:

And also, here is an even simpler example which shows this undesired behavior.

#!/usr/bin/env python
# coding: utf-8

# ### Differentiating an implicit function using SymPy

# In[1]:


import sympy as sp


# In[2]:


sp.__version__


# In[3]:


sp.init_printing(use_latex='mathjax')  # use pretty mathjax output


# In[4]:


sp.var('x y')

F = sp.ln(sp.sqrt(x**2 + y**2)) - sp.atan(y / x)


# In[5]:


f1 = sp.idiff( F, y, x ) # First derivative of y w.r.t. x
f1


# In[6]:


sp.simplify(f1)


# In[7]:


f2 = sp.idiff( F, y, x, 2) # Second derivative of y w.r.t. x
f2

sp.simplify(f2)


# In[ ]:

The second derivative here is given as:

This expression obviously can be simplified further even without using any special facts.

Answer 1

You can simplify the expressions yourself. In the first example you can just choose a term to eliminate and solve F for that:

In [42]: F
Out[42]: 
     sin(y)    sin(x)
- x⋅ℯ       + ℯ      

In [43]: solve(F, exp(sin(y)))
Out[43]: 
⎡ sin(x)⎤
⎢ℯ      ⎥
⎢───────⎥
⎣   x   ⎦

In [44]: [esy] = solve(F, exp(sin(y)))

In [45]: f2.subs(exp(sin(y)), esy)
Out[45]: 
                        2                                     
            x⋅sin(y)⋅cos (x)   2⋅sin(y)⋅cos(x)     sin(y)    1
-x⋅sin(x) + ──────────────── - ─────────────── + ───────── + ─
                   2                  2               2      x
                cos (y)            cos (y)       x⋅cos (y)    
──────────────────────────────────────────────────────────────
                           x⋅cos(y) 

You can apply further simplification operations from there.

In the second example you can just call factor :

In [47]: f2
Out[47]: 
         ⎛ 2    2⎞       
       2⋅⎝x  + y ⎠       
─────────────────────────
 3      2          2    3
x  - 3⋅x ⋅y + 3⋅x⋅y  - y 

In [48]: factor(f2)
Out[48]: 
  ⎛ 2    2⎞
2⋅⎝x  + y ⎠
───────────
         3 
  (x - y)