YES

Problem:
 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))

Proof:
 DP Processor:
  DPs:
   D#(+(x,y)) -> D#(y)
   D#(+(x,y)) -> D#(x)
   D#(*(x,y)) -> D#(y)
   D#(*(x,y)) -> D#(x)
   D#(-(x,y)) -> D#(y)
   D#(-(x,y)) -> D#(x)
  TRS:
   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))
  LPO Processor:
   argument filtering:
    pi(t) = []
    pi(D) = [0]
    pi(1) = []
    pi(constant) = []
    pi(0) = []
    pi(+) = [0,1]
    pi(*) = [0,1]
    pi(-) = [0,1]
    pi(D#) = 0
   precedence:
    D > D# ~ - ~ * ~ + ~ 0 ~ constant ~ 1 ~ t
   problem:
    DPs:
     
    TRS:
     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))
   Qed