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.
Conjugate
isn't a differentiable function. $\endgroup$t
andsqr
are to be treated as real, tryD[ComplexExpand[expr], t]
. Variables in Mma are treated as complex by default. $\endgroup$