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))
 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)))

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)
   D#(minus(x)) -> D#(x)
   D#(div(x,y)) -> D#(y)
   D#(div(x,y)) -> D#(x)
   D#(ln(x)) -> D#(x)
   D#(pow(x,y)) -> D#(y)
   D#(pow(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))
   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)))
  Usable Rule 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)
    D#(minus(x)) -> D#(x)
    D#(div(x,y)) -> D#(y)
    D#(div(x,y)) -> D#(x)
    D#(ln(x)) -> D#(x)
    D#(pow(x,y)) -> D#(y)
    D#(pow(x,y)) -> D#(x)
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [D#](x0) = x0,
     
     [ln](x0) = 1x0,
     
     [pow](x0, x1) = 1x0 + 2x1 + 3,
     
     [div](x0, x1) = 1x0 + 1x1 + 3,
     
     [minus](x0) = 4x0 + 0,
     
     [-](x0, x1) = 1x0 + 4x1 + 0,
     
     [*](x0, x1) = 1x0 + 1x1 + 6,
     
     [+](x0, x1) = 1x0 + 2x1 + 1
    orientation:
     D#(+(x,y)) = 1x + 2y + 1 >= y = D#(y)
     
     D#(+(x,y)) = 1x + 2y + 1 >= x = D#(x)
     
     D#(*(x,y)) = 1x + 1y + 6 >= y = D#(y)
     
     D#(*(x,y)) = 1x + 1y + 6 >= x = D#(x)
     
     D#(-(x,y)) = 1x + 4y + 0 >= y = D#(y)
     
     D#(-(x,y)) = 1x + 4y + 0 >= x = D#(x)
     
     D#(minus(x)) = 4x + 0 >= x = D#(x)
     
     D#(div(x,y)) = 1x + 1y + 3 >= y = D#(y)
     
     D#(div(x,y)) = 1x + 1y + 3 >= x = D#(x)
     
     D#(ln(x)) = 1x >= x = D#(x)
     
     D#(pow(x,y)) = 1x + 2y + 3 >= y = D#(y)
     
     D#(pow(x,y)) = 1x + 2y + 3 >= x = D#(x)
    problem:
     DPs:
      
     TRS:
      
    Qed