Error when using scalar vs. vector to operate on tracked inupt

Hi ReverseDiff team,

I've found an error that comes up when an object used in a function being differentiated is the result of a tracked array and another argument (not the one we are finding the derivative with respect to) when the other argument is a scalar. I understand my explanation is a bit imprecise, a minimum reproducible example is below. In the line labeled "doesn't work" the arg c is treated as a scalar. In the line labeled "works" it is treated as a length one vector. The object d is used in further computation; when c is treated as a scalar this leads to a method error when it (an element of the resulting object d) is used in the line aa*sin(dd).

[EDIT]: I'm on Julia 1.8.3 and the version of ReverseDiff.jl on the Julia general repo (v1.14.4).

using ReverseDiff
using Distributions

# a,b are vectors; want Hessian w.r.t a
# c is "scalar" arg
function mwe(a,b,c)
    d = @. a/(b+c)
    e = map(a,b,d) do aa, bb, dd
        if dd < bb
            exp(bb)
        else
            aa*sin(dd)
        end
    end
    return sum(e)
end

a = rand(Normal(),5)
b = rand(Normal(),5)
c = 50.0

ReverseDiff.hessian(x -> mwe(x, b, c), a)  # doesn't work
ReverseDiff.hessian(x -> mwe(x, b, [c]), a)  # works

Start by running the provided mwe function with ReverseDiff.hessian on Julia 1.8.3 and ReverseDiff.jl v1.14.4, comparing scalar c with one-element vector [c]. Trace the failing path through the scalar case and verify that the Hessian completes without the method error seen at aa*sin(dd).