YES
We show the termination of the TRS R:

  D(t()) -> |1|()
  D(constant()) -> |0|()
  D(+(x,y)) -> +(D(x),D(y))
  D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
  D(-(x,y)) -> -(D(x),D(y))
  D(minus(x)) -> minus(D(x))
  D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
  D(ln(x)) -> div(D(x),x)
  D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))


-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: D#(+(x,y)) -> D#(x)
p2: D#(+(x,y)) -> D#(y)
p3: D#(*(x,y)) -> D#(x)
p4: D#(*(x,y)) -> D#(y)
p5: D#(-(x,y)) -> D#(x)
p6: D#(-(x,y)) -> D#(y)
p7: D#(minus(x)) -> D#(x)
p8: D#(div(x,y)) -> D#(x)
p9: D#(div(x,y)) -> D#(y)
p10: D#(ln(x)) -> D#(x)
p11: D#(pow(x,y)) -> D#(x)
p12: D#(pow(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: D#(+(x,y)) -> D#(x)
p2: D#(pow(x,y)) -> D#(y)
p3: D#(pow(x,y)) -> D#(x)
p4: D#(ln(x)) -> D#(x)
p5: D#(div(x,y)) -> D#(y)
p6: D#(div(x,y)) -> D#(x)
p7: D#(minus(x)) -> D#(x)
p8: D#(-(x,y)) -> D#(y)
p9: D#(-(x,y)) -> D#(x)
p10: D#(*(x,y)) -> D#(y)
p11: D#(*(x,y)) -> D#(x)
p12: D#(+(x,y)) -> D#(y)

and R consists of:

r1: D(t()) -> |1|()
r2: D(constant()) -> |0|()
r3: D(+(x,y)) -> +(D(x),D(y))
r4: D(*(x,y)) -> +(*(y,D(x)),*(x,D(y)))
r5: D(-(x,y)) -> -(D(x),D(y))
r6: D(minus(x)) -> minus(D(x))
r7: D(div(x,y)) -> -(div(D(x),y),div(*(x,D(y)),pow(y,|2|())))
r8: D(ln(x)) -> div(D(x),x)
r9: D(pow(x,y)) -> +(*(*(y,pow(x,-(y,|1|()))),D(x)),*(*(pow(x,y),ln(x)),D(y)))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            D#_A(x1) = max{0, x1 - 236}
            +_A(x1,x2) = max{74, x1 + 14, x2 + 14}
            pow_A(x1,x2) = max{72, x1 + 14, x2 + 17}
            ln_A(x1) = max{74, x1 + 7}
            div_A(x1,x2) = max{x1 + 235, x2 + 175}
            minus_A(x1) = max{74, x1 + 14}
            -_A(x1,x2) = max{74, x1 + 14, x2 + 14}
            *_A(x1,x2) = max{x1 + 14, x2 + 79}
            D#!_A(x1) = max{3, x1 - 3}
            +!_A(x1,x2) = 4
            pow!_A(x1,x2) = max{76, x1 - 2, x2 + 1}
            ln!_A(x1) = max{77, x1 - 3}
            div!_A(x1,x2) = max{161, x1 - 52, x2 + 28}
            minus!_A(x1) = 100
            -!_A(x1,x2) = max{x1 + 14, x2 - 300}
            *!_A(x1,x2) = x2 + 161
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              D#_A(x1) = 0
              +_A(x1,x2) = 0
              pow_A(x1,x2) = x1 + 1
              ln_A(x1) = 1
              div_A(x1,x2) = 3
              minus_A(x1) = 1
              -_A(x1,x2) = 5
              *_A(x1,x2) = x2 + 1
              D#!_A(x1) = 1
              +!_A(x1,x2) = 4
              pow!_A(x1,x2) = 2
              ln!_A(x1) = x1 + 6
              div!_A(x1,x2) = 4
              minus!_A(x1) = 0
              -!_A(x1,x2) = x1 + 6
              *!_A(x1,x2) = 3
        
  
    partial status:
  
      pi(D#) = []
      pi(+) = [1]
      pi(pow) = [2]
      pi(ln) = [1]
      pi(div) = [1, 2]
      pi(minus) = [1]
      pi(-) = [1]
      pi(*) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12

Then no dependency pair remains.