Mathematica: take derivative in terms of original function

[PlasticOh-No]

Hello,

Let's say we have

f[x_,y_]:= whatever

then we say

expr1 = D[f[x,y],x]

which is some complicated expression.

How can we get the answer in terms of f[x,y]? That is, how can we get expr1 to look like:

expr1 = f[x,y]*this term + pi*Sqrt[f[x,y]] + et cetera?

Thanks Gurus

 

[PlasticOh-No]

Ok let me make this a little clearer.
Let's say we have
f[x_,y_]:=Exp[a*(x+y)]
where a is some real parameter.
Now we take
D[f[x,y],x]
which is equal to
a*Exp[a*(x+y)].
How can I get Mathematica to give a result like
a*f[x,y]

Thank you in advance for you attention on this.

 

[Bill Simpson]

Try this and see if it will do what you want.
After you have taken the derivative then Undefine f[x,y] and then try
yourResult/.Exp[a*(x+y)]->f[x,y]
and see if you can get it to backsubstitute into the form you desire.

The reason you need to undefine f[] is to keep Mathematica from immediately expanding back to the definition.

This is a somewhat error prone method, but if used with care and checking can often work.

 

[PlasticOh-No]

Thanks for your reply. Just one thing - how do I undefine the original definition of f[x_,y_] ??

Thanks again

 

[Bill Simpson]

http://reference.wolfram.com/mathematica/ref/Clear.html

 

[PlasticOh-No]

Thank you