MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Chain rule


Once you eveluate x=2t +5 in Mathematica it no longer makes sense to
differentiate anything with respec to x:

In[27]:=
x = 2t + 1

Out[27]=
1 + 2 t

In[28]:=
D[f, x]

General::"ivar": "1 + 2t is not a valid variable."

Out[28]=
D[f, 1 + 2 t]

However, you can still make Mathematica "use the Chain rule" in the way you
want by making use of Mathematica's ability to solve equations, and
eliminate variables. Here is an example using Solve (one can also use
GroebnerBasis or Reduce):

In[29]:=
eq1 = x - 2t - 5; eq2 = y - 2t^2 + 1;


In[26]:=
Solve[{eq1 == 0, eq2 == 0, Dt[eq1, x] == 0, Dt[eq2, x] == 0}, {Dt[y, x],
    x}, {Dt[t, x], y, t}]
Solve::"svars": "Equations may not give solutions for all \"solve\" \
variables."
Out[26]=
{{Dt[y, x] -> -5 + x}}

Note that you have to use the total differential Dt rather than the
derivative D.



-- 
Andrzej Kozlowski
Toyama International University, JAPAN

For Mathematica related links and resources try:
<http://www.sstreams.com/Mathematica/>



on 7/24/00 9:04 AM, Claude at fran_claude at hotmail.com wrote:

> Hello,
> 
> I am just trying to solve:
> dy/dx when x = 2t + 5 and y = 2t^2 - 1.
> 
> Can it be done directly by Mathematica or do I have to solve dy/dt and
> dx/dt and finally divide dy/dt by dx/dt?
> 
> I tried posing x = 2t + 5 and y = 2t^2 - 1 and ask solve for D [y,x], but
> that does not evaluate properly (the answer is 2t).  I would have thought
> that Mathematica would have used the chain rule automatically.
> 
> Thanks for your help,
> 
> 
> 
> 



  • Prev by Date: continuously updating mean confidence interval (MeanCI)
  • Next by Date: Re: Equation of a "potato"
  • Previous by thread: Re: Chain rule
  • Next by thread: Mathematica 3.0 errors