ReplaceAll does not work on derivatives

Question

(q''[x] - Exp[q[x]/2)] /. q[x] -> y[x] + a*t[x]

gives

-Exp^[1/2 (y[x] + a*t[x])] + q''[x]

Which is not what I want.

I would like to have

-Exp^[1/2 (y[x] + a*t[x])] + y''[x] + a*t''[x]

Is that possible?

Do this: q -> (y[#] + a*t[#]&). (Duplicate, but I don't have time to look for it.) — march, Commented Oct 28, 2016 at 15:32
Nice, the search engine didn't show me anything, but I am bad at formulating questions. — ThunderBiggi, Commented Oct 28, 2016 at 15:38

Szabolcs · Accepted Answer · 2016-10-28 15:45:04Z

6

Check the FullForm of your expression. That will make it clear why it doesn't work.

The solution is to replace q and not q[x].

q''[x] - Exp[q[x]/2] /. q -> Function[x, y[x] + a*t[x]]

Recommended reading:

answered Oct 28, 2016 at 15:33

Szabolcs

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$\begingroup$ I checked FullForm before posting and it showed me Derivative[2][q][x], hence I assumed (without checking) that if I replace just q with a function of x I will get something like (...[x])[x] and so I didn't even try it $\endgroup$
– ThunderBiggi
Commented Oct 28, 2016 at 15:40
1

$\begingroup$ @user1482714 q -> f[x] of course won't work. You need q -> Function[x, f[x]]. $\endgroup$
– Szabolcs
Commented Oct 28, 2016 at 15:44

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