Derivative with respect to a variable is leaving unevaluated derivatives of functions

Question

I have the following

Inactive[Log][Sqrt[
 E^(-0.96 t - 0.96 Conjugate[t])
   Conjugate[
   Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/sqr] (Cosh[(sqr t)/2] - (
    1.76 Sinh[(sqr t)/2])/sqr)]]

Removing inactivate, upon taking the derivative w.r.t t, I end up with

E^(0.96 t + 
    0.96 Conjugate[t]) (E^(-0.96 t - 0.96 Conjugate[t])
       Conjugate[
       Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/
        sqr] (-0.88 Cosh[(sqr t)/2] + 1/2 sqr Sinh[(sqr t)/2]) + 
     E^(-0.96 t - 0.96 Conjugate[t])
       Conjugate[
       Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/
        sqr] (Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/sqr) (-0.96 - 
        0.96 Derivative[1][Conjugate][t]) + 
     E^(-0.96 t - 
       0.96 Conjugate[t]) (Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/
        sqr) (-0.88 Cosh[(sqr t)/2] + 
        1/2 sqr Sinh[(sqr t)/2]) Derivative[1][Conjugate][
       Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/sqr]))/(4 Conjugate[
    Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/sqr] (Cosh[(sqr t)/2] - (
     1.76 Sinh[(sqr t)/2])/sqr)

This gives me 54.1743 (-0.0204715 - 0.0204715 Derivative[1][Conjugate][0.177417] + 0.00461473 (-0.96 - 0.96 Derivative[1][Conjugate][1])) when a value for t is passed in. As you can see, the issue is the unevaluated D[conjugate[t],t], which prevents me from obtaining a plottable value. I have attempted to make use of Inactivate and activate, as other posts with a similar issue have mentioned, but this doesn't seem to be addressing the issue.

As I see it, I need all derivatives w.r.t t to be evaluated before the replace all comes into play for any value of t, which I thought inactivate and activate would allow. But this does not seem to be the case, as I just get the value of t is not a variable, which makes sense given its trying then to take the derivative with respect to a number. What am I missing here?

Edit: I will also say that there are, obviously, a few more variable passed in here that are not mentioned, but since they are not causing an issue with the derivation, I have omitted them from the code.

Answer 1

If expr is your expression with the Inactivate, and you're looking for a complex (Wirtinger) derivative, then we can use ResourceFunction["ComplexD"] and simplify assuming t is real valued:

FullSimplify[
 ResourceFunction["ComplexD"][expr, t]
 , t ∈ Reals]

Which gives us:

(-0.92 sqr Cosh[(sqr t)/2] + (0.8448 + 0.25 sqr^2) Sinh[(sqr t)/2])/(
sqr Cosh[(sqr t)/2] - 1.76 Sinh[(sqr t)/2])

If you use RealD from that page:

RealD[f_, z_] := 
 ResourceFunction["ComplexD"][f, z] + 
  ResourceFunction["ComplexD"][f, Conjugate[z]]

Then we get:

FullSimplify[
 RealD[expr, t]
 , t ∈ Reals]

Result:

(-1.4 sqr Cosh[(sqr t)/
     2] + (Conjugate[-0.88 Cosh[(sqr t)/2] + 
        0.5 sqr Sinh[(sqr t)/2]] (0.5 sqr Cosh[(sqr t)/2] - 
        0.88 Sinh[(sqr t)/2]))/
    Conjugate[
     Cosh[(sqr t)/2] - (1.76 Sinh[(sqr t)/2])/sqr] + (1.6896 + 
      0.25 sqr^2) Sinh[(sqr t)/2])/(sqr Cosh[(sqr t)/2] - 
   1.76 Sinh[(sqr t)/2])

If you don't care about Conjugate then you could just do:

FullSimplify[D[expr /. Conjugate -> Identity, t]]