# Display functions without dependence?

I'm solving systems of equations with a few different functions of time, t, and so I'm getting output with lots of [t] running around. Is there a way to display these functions without the dependences, e.g., to display

a[t] b'[t]


as

a b'


? I can use a series of replacement rules to do this, e.g.,

/.{a[t]->a,b[t]->b}


but this is pretty cumbersome, and doesn't generalize well (e.g., when more functions are added in).

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Something like a[t] b'[t] /. f_[t] :> f? Please note that this could fail, depending on your expression. –  Yves Klett Jan 10 '14 at 12:22
Hi Yves, that works, thanks! Why do you use the delayed :> instead of ->? –  Adam Jan 10 '14 at 12:30
Right, Rule (-> ) works just as well in your case :) –  Yves Klett Jan 10 '14 at 12:44
@Yves With respect, no, it does not. You should use :> with named patterns on the right-hand-side. Otherwise if f has a global value the replacement will break. –  Mr.Wizard Jan 10 '14 at 12:49
This rule does not work for all functions of the independent variable. For example a[t] Exp[t] gives a Exp[t]. But may be that is ok with the OP, just wanted to point it out just in case. –  Nasser Jan 10 '14 at 12:51