2007-04-30 16:23:02 · 4 answers · asked by Superconductor 3 in Science & Mathematics ➔ Mathematics
w/respect to u..
Why do u take derivative of cos u, instead of cos^3 u as in
3cos^2u d/du (cos u)
2007-04-30 16:31:15 · update #1
v = cos^3(u)
v ' = 3cos^2 (u) x (-sin(u))
v ' = -3sin(u)cos^2 (u)
2007-04-30 16:29:45 · answer #1 · answered by frank 7 · 0⤊ 0⤋
v(u) = (cos u)³
Next step in crude terms is "differentiate bracket and then differentiate what is inside bracket" :-
v`(u) = 3.(cos u)² (- sin u)
v` (u) = - 3.cos² u.sin u
2007-05-01 05:37:19 · answer #2 · answered by Como 7 · 0⤊ 0⤋
Assuming the derivative is w.r.t u,
dv/du = 3cos^2(u). d(cosu)/du
= 3cos^2(u) (-sinu)
= - 3cos^2(u)sin(u)
2007-04-30 23:28:38 · answer #3 · answered by Sumeer 1 · 0⤊ 0⤋
Use the chain rule.
2007-04-30 23:28:22 · answer #4 · answered by Paul W 2 · 0⤊ 0⤋