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If I have an equation
eq = f[x] + g[x] == 0
and I solve for f[x]
sol = Solve[eq,f[x]]
how do I use sol to replace both f[x] and its derivatives in another equation? For example
eq2 = f'[x]*g[x] + f[x] == 0
and then
eq2 /. sol
outputs
{-g[x] + g[x] f'[x] == 0}
The problem is caused by the internal structure of Derivative:
f'[x] // FullForm
(* ==> Derivative[1][f][x] *)
There's no f[x] in there to replace. Using D[f[x], x] instead of f'[x] and Holding the whole equation works:
eq2 = Hold[D[f[x], x]*g[x] + f[x] == 0];
ReleaseHold[eq2 /. sol]
(* ==> {-g[x] - g[x] Derivative[1][g][x] == 0} *)
eq2 b/c I need to do some other calculations on it first. So I tried ReleaseHold[Hold[eq2 /. f'[x] -> D[f[x], x]] /. sol] which gives {f[x] - g[x] g'[x]}. But ReleaseHold[Hold[eq2 /. f'[x] -> D[f[x], x]] /. sol] /. sol does give the desired answer.
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If you plan to use derivatives, or solve for functions, you could use DSolve instead of Solve.
sol = DSolve[eq, f, x];
eq2 /. sol
{-g[x] - g[x] Derivative[1][g][x] == 0}
If not, apart from avoiding the use of x overall, you could do something like this
parseRule = (f_[x_] -> sth_) :> (f -> (Evaluate[sth /. x -> #] &));
sol = Solve[eq, f[x]]
eq2 /. (sol /. parseRule)
{-g[x] - g[x] Derivative[1][g][x] == 0}
I have the feeling there are better options. We'll see
An approach that may in the long run help avoid confusion in your calculation is to use a different name for the function that is the solution to the equation eq = f[x] + g[x] == 0.
What I mean is this:
fSolved[x_] = f[x] /. First@Solve[eq, f[x]]
eq2 = fSolved'[x]*g[x] + fSolved[x] == 0
(* ==> {-g[x] - g[x] Derivative[1][g][x]} == 0 *)
So here I chose the name fSolved for the actual solution with which I want to work later.
This is mathematically more sane, I think, because Solve for eq could in principle give you several possible solutions in the form of a list of rules f[x] -> ..., and then you'd have to give each solution different names anyway if you want to keep working with both of them. As an example, consider
eq = f[x]^2 - g[x] == 0
Then by using First above we select the negative square root, and we could use a different name for the positive square root (Last@Solve...).
:=Evaluate instead of plain = ? Syntax coloring?
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fSolved[y] in eq2 (i.e., be independent of the name x for the variable) if we feel like it...
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fSolved[x_]:=Evaluate[...] instead of fSolved[x_]=...
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Evaluate.
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