2006-12-03 19:49:16 · 5 answers · asked by car_lover_42020 2 in Science & Mathematics ➔ Mathematics
d/dx [(15-x)/3] = d/dx (15/3) - d/dx (x/3)
= d/dx 5 - d/dx (x/3)
= 0 - 1/3
= -1/3
2006-12-03 20:00:36 · answer #1 · answered by the DoEr 3 · 0⤊ 0⤋
Simple
d/dx[(15-x)/3]=1/3 [d/dx(15)] + 1/3 [d/dx(-x)]
= 1/3 * 0 + 1/3 * (-1)
= 0 + (-1/3)
= -1/3
2006-12-04 03:59:48 · answer #2 · answered by Cool Sun 3 · 0⤊ 0⤋
d/dx [(15-x)/3] = d/dx (15/3) - d/dx (x/3)
= d/dx 5 - d/dx (x/3)
= 0 - 1/3
= -1/3
2006-12-04 05:58:44 · answer #3 · answered by jethro_the_retard 2 · 0⤊ 0⤋
(15-x)/3 = 5-1/3*x
d/dx[(15-x)/3]
=d/dx[5-1/3*x]
=d/dx[5] - d/dx[1/3*x]
=0-1/3
=-1/3
2006-12-04 05:10:45 · answer #4 · answered by Ranna Renni 2 · 0⤊ 0⤋
-1/3
Doug
2006-12-04 03:51:52 · answer #5 · answered by doug_donaghue 7 · 0⤊ 0⤋