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MathGroup Archive 1998

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Re: Derivative via mathematica



cai wrote:
> 
> Hi,
> 
> I just used mathematica for a couple of days.  I am trying to compute
> the derivative under mathematica.  Because the function is complicated,
> I like to break it down.
> 
> f[t_] = (m/(1+Exp[1/t] +b)
> 
> Here m and b are functions of t.
> If I directly use command D after insert m and b terms, a very
> complicated equaion is gerenated, which I do not want.
> 
> What I want is if I define the values of m' and b', rewrite the f
> 
> m' = p
> b' = q   // well, I dont know how to define, this is the idea
> 
> f[t_, m[t], b[t]] = (m/(1+Exp[1/t] +b)
> 
> then use the command D[f[t,m[t],b[t]],t] hopeful get a equation which is
> the function of t, p and q.  How to do that?
> 
> A related question, I tried to use non-defined function In[19]:= m[t_]
> Out[19]= m[t_]
> In[20]:= D[m[t],t]
> Out[20]= m'[t]
> and expected D[f[t,m[t],b[t]],t] contains m'[t].  Is it possible?
> 
> Basically, it is a chain derivative question, I just want it to stop
> earlier.
> 
> Could somebody help?
> Thanks.
> 
> ccai1@ohiou.edu
In[25]:=
Clear["`*"]

In[26]:=
$Line = 0;

Wan, here are some suggestions and comments. They do not address all of
your points but I hope that they may be of help - please mail me if you
need more help.

Mathematica, of course, uses the standard results

In[1]:=
D[a f[t]+ b g[t],t]

Out[1]=
a f'[t] + b g'[t]

In[2]:=
D[f[t] g[t],t]

Out[2]=
g[t] f'[t] + f[t] g'[t]

In[3]:=
D[1/ g[t],t]

Out[3]=
  g'[t]
-(-----)
      2
  g[t]

In[4]:=
 D[f[g[t]],t]

Out[4]=
f'[g[t]] g'[t]

And we can put in values for f and g ,  f' and g' retrospectively. This
will be demonstrated in the treatment of the example below

In[5]:=
f1[t_] = m/(1+Exp[1/t] +b)

Out[5]=
     m
------------
         1/t
1 + b + E

Mathematica assumes that m and b are independent of t.


In[6]:=
d1=D[f1[t],t]

Out[6]=
       1/t
      E    m
------------------
          1/t 2  2
(1 + b + E   )  t

Which is not what you want.
We can avoid this by using

In[7]:=
f2[t_] = m[t]/(1+Exp[1/t] +b[t])

Out[7]=
     m[t]
---------------
     1/t
1 + E    + b[t]

Now we get,

In[8]:=
d2=D[f2[t],t]

Out[8]=
           1/t
          E
  m[t] (-(----) + b'[t])
            2
           t                     m'[t] -(----------------------) +
---------------
          1/t        2           1/t
    (1 + E    + b[t])       1 + E    + b[t]

If we  define the functions m and b

In[9]:=
m[t_] := Sin[t]; b[t_]:= Exp[k t]

we get

In[10]:=
d2

Out[10]=
                             1/t
                    k t     E
                  (E    k - ----) Sin[t]
                              2
    Cos[t]                   t
--------------- - ----------------------
     1/t    k t           1/t    k t 2 1 + E    + E        (1 + E    + E
)

It is possible to define the derivatives of these functions but we
cannot consistently define a function ad its derivative independently
so we clear the
definitions above (there are also programming reasons for this)

In[11]:=
Clear[m,b]

In[12]:=
m'[t_] = Cos[t^2]; b'[t_]:= t Exp[k t];

These derivative definitiions are used

In[13]:=
d2

Out[13]=
                      1/t
                     E        k t
                  (-(----) + E    t) m[t]
         2             2
    Cos[t ]           t
--------------- - -----------------------
     1/t                  1/t        2 1 + E    + b[t]     (1 + E    +
b[t])

but Mathematica has left m[t]; it has not integrated m'[t] to find it.
Defining derivatives can be useful but perhaps not so much here.
Incidentally the derivatives are stored under Derivative, not m and b.

In[14]:=
??m

>From In[14]:=
Global`m

In[15]:=
??Derivative

>From In[15]:=
f' represents the derivative of a function f of one argument.
   Derivative[n1, n2, ... ][f] is the general form,
   representing a function obtained from f by differentiating
   n1 times with respect to the first argument, n2 times with
   respect to the second argument, and so on.
Derivative[1][m][t_] = Cos[t^2]
 
Derivative[1][b][t_] := t*Exp[k*t]

This, and the notation m'[t], b'[t], leads us to the important contrast
between functions and formulas: 
g' is the first derivative of the function g;  g'[t] is its value at t -
it is
the same as D[g[t],t], the derivative of the formula g[t] with respect
to the variable t.
The FullForm of g' is

In[16]:=
FullForm[g']

Out[16]//FullForm=
Derivative[1][g]

In[17]:=
g'[t]

Out[17]=
g'[t]

In[18]:=
D[g[t],t]

Out[18]=
g'[t]

Once you get into pure functions you may wish to use a different
apporoach which reduces the chance of interference between definitions.
But first, I need to clear the definitions of the derivatives.

In[19]:=
Clear[Derivative]


In[20]:=
f2[t]/.{ m -> Function[t, Sin[t]],
                b -> Function[t, Exp[k t]]}

Out[20]=
    Sin[t]
---------------
     1/t    k t
1 + E    + E

In[21]:=
d2/.{ m ->Sin,
                b -> Function[t, Exp[k t]]}

Out[21]=
                             1/t
                    k t     E
                  (E    k - ----) Sin[t]
                              2
    Cos[t]                   t
--------------- - ----------------------
     1/t    k t           1/t    k t 2 1 + E    + E        (1 + E    + E
)

or, more briefly,

In[22]:=
d2/.{ m -> Sin,b -> (Exp[k #]&)}

Out[22]=
                             1/t
                    k t     E
                  (E    k - ----) Sin[t]
                              2
    Cos[t]                   t
--------------- - ----------------------
     1/t    k t           1/t    k t 2 1 + E    + E        (1 + E    + E
)


A variant of your idea for f can be implemented by

In[23]:=
f[m_,b_]= f2[t]

Out[23]=
     m[t]
---------------
     1/t
1 + E    + b[t]

In[24]:=
D[f[ Cos, Exp[k #]&],t]

Out[24]=
             1/t
    k t     E
  (E    k - ----) Cos[t]
              2
             t                  Sin[t] -(----------------------) -
---------------
          1/t    k t 2           1/t    k t
    (1 + E    + E   )       1 + E    + E

You might like to look up the following in the Help Browser D
Derivative
Dt  (total derivative)

Please notice that these functions also deal with several variables. -- 
-- 
Allan Hayes
Training and Consulting
Leicester, UK
hay@haystack.demon.co.uk
http://www.haystack.demon.co.uk
voice: +44 (0)116 271 4198
fax: +44 (0)116 271 8642




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