2007-12-28 01:23:45 · 5 answers · asked by mix 1 in Science & Mathematics ➔ Mathematics
y = cos 2x + sin 2x
dy/dx = - 2 sin 2x + 2 cos 2x
2007-12-28 01:50:21 · answer #1 · answered by Como 7 · 3⤊ 0⤋
d/dx (cos2x + sin2x)
= d(cos2x)/dx + d(sin2x)/dx
= (-sin2x)2 + (cos2x)2
= -2sin2x + 2(cos2x)
= 2[-sin2x + cos2x] ANS
teddy boy
2007-12-28 09:48:27 · answer #2 · answered by teddy boy 6 · 0⤊ 0⤋
d/dx (cos2x + sin2x)
= d/dx (cos 2x) + d/dx (sin 2x)
= - sin 2x * d/dx (2x) + cos 2x * d/dx (2x) (chain-rule)
= - 2 sin 2x + 2 cos 2x
2007-12-28 09:30:09 · answer #3 · answered by Madhukar 7 · 0⤊ 0⤋
d/dx(cos2X+sin2X)= -2sin2X+2cos2X
2007-12-28 09:28:04 · answer #4 · answered by Tarun S 2 · 0⤊ 0⤋
d/dx (cos2x + sin2x)
= [d/dx (cos 2x)] + [d/dx (sin2x)]
= [(-sin2x) . d/dx (2x) ] + [ (cos 2x) d/dx (2x) ]
= [ (-sin2x) . (2) ] + [ (cos 2x) . (2) ]
= [ - 2 sin 2x ] + [ 2 cos 2x ]
= 2 [ cos 2x - sin 2x ] ........................... Answer
2007-12-28 10:46:33 · answer #5 · answered by Pramod Kumar 7 · 0⤊ 1⤋