2006-12-19 16:19:33 · 13 answers · asked by abhay k 1 in Science & Mathematics ➔ Mathematics
y'=xcosx+sinx+sinx+xcosx
=2(sinx+xcosx)
aliter
y=2xsinx
y'=2[xcosx+sinx]
2006-12-19 16:22:06 · answer #1 · answered by raj 7 · 0⤊ 0⤋
y = 2 x sin x
dy/dx =2 (x cos x + sinx )
=2x cos x +2 sin x
2006-12-19 18:19:00 · answer #2 · answered by zido 1 · 0⤊ 0⤋
y=2(x*sin x)
dy/dx=2(x*cos x+sin x)
2006-12-19 20:21:42 · answer #3 · answered by cham 2 · 0⤊ 0⤋
7
2006-12-19 16:26:56 · answer #4 · answered by mernieinc 4 · 0⤊ 3⤋
y = (x)sin x + (sin x)x
dy/dx=xcosx+sinx+(sin x)x(differentiating wrt to x)
dy/dx =xcosx+sinx+xcosx+sinx
using the uv rule, derivtive of uv = udv/dx+v du/dx,(x)sin x=xcosx+sinx and similarly for (sin x)x
so
dy/dx=2(sinx+xcosx)
hope this helps..
2006-12-19 16:45:40 · answer #5 · answered by For peace 3 · 0⤊ 0⤋
d/dx x(sin x) = x(cos x) + sin x
d/dx (sin x)x = sin x + (cos x)x
dy/dx = 2 [x(cos x) + sin x]
2006-12-19 16:30:23 · answer #6 · answered by Johnny Handsome 2 · 0⤊ 0⤋
y = x(sinx)+(sinx)x
= 2x sinx
dy/dx = 2 d(xsinx)/dx
= 2 xcosx+sinx ( product formula)
2006-12-19 20:22:42 · answer #7 · answered by george t 2 · 0⤊ 0⤋
y = (x)sin x + (sin x)x
y = (2x)sin x
dy/dx = 2sin x + (2x)cos x = 2[sin x + (x)cos x]
2006-12-19 17:34:15 · answer #8 · answered by Northstar 7 · 0⤊ 0⤋
y = (x)sin x + (sin x)x
y = 2(x)sin x
y' = 2(x cosx + sinx)
2006-12-19 16:44:30 · answer #9 · answered by Helmut 7 · 0⤊ 0⤋
hi the conception is to transpose the formula as shown in college. that's what the practice is when you doing the artwork not replying on some different person to do the transposition for you.
2016-11-30 23:54:03 · answer #10 · answered by ? 4 · 0⤊ 0⤋
dy/dx = sin(x) + xcos(x) + sin(x) + xcos(x)
dy/dx = 2sin(x) + 2xcos(x)
2006-12-19 16:52:32 · answer #11 · answered by Mr.Math 1 · 0⤊ 0⤋