Question

need help on1st-4th derivatives of sin(x-x^2).
currently stucked at 1st derivative, obtained:

(sin x sin x^2)(1-2x) + (cos x cos x^2)(1-2x),

but the calculator showing the 1st derivative to be:

(1-2x) cos[(x-1)x]

Answer 1

why you using a calculator?

Answer 2

to check my 1st derivative

Answer 3

its an online calculator @_@ LOL

Answer 4

what D(sin) ?

Answer 5

cos?

Answer 6

yes, soo

cos(x-x^2) * D(x-x^2)

what D(x-x^2?)

Answer 7

i opened sin(x-x^2) into sin x cos x^2 - cos x sin x^2
then i use product rule on both o.o

or im not supposed to open it?

Answer 8

ummm, i wouldnt try to invent new ways of failing just yet ...

Answer 9

oooh...wait uh

Answer 10

its not needed to go thru all that stuf an create alot of work for yourself

Answer 11

D sin(x-x^2) = (1-2x) cos (x-x^2) ?

Answer 12

D(sin(f(x))) = cos(f(x)) * D(f(x))

thats the chain rule

Answer 13

yes

Answer 14

now you can distribute the cos(x-x^2) thru or leave it as is and continue to the D2

Answer 15

my initial working straight continued into D2, got (1-2x)^2 [-sin(x-x^2)] -2cos(x-x^2)

Answer 16

D (1-2x) cos (x-x^2) = D(1-2x) cos (x-x^2) + (1-2x) Dcos (x-x^2)

Answer 17

-2cos (x-x^2) - (1-2x)^2 sin (x-x^2)

Answer 18

so thats good, D3 it now

Answer 19

-(1-2x)^3 cos(x-x^2) +4(1-2x) sin(x-x^2) +2(1-2x) sin(x-x^2) - cos(x-x^2) o.o

Answer 20

D -2cos (x-x^2) - [ D(1-2x)^2 sin (x-x^2) + (1-2x)^2 D sin (x-x^2) ]

\[2(1-2x) sin (x-x^2) - [ -4(1-2x) sin (x-x^2) + (1-2x)^3 cos (x-x^2) ]\]
\[2(1-2x) sin (x-x^2) + 4(1-2x) sin (x-x^2) - (1-2x)^3 cos (x-x^2)\]
\[(2-4x) sin (x-x^2) + (4-8x) sin (x-x^2) - (1-2x)^3 cos (x-x^2)\]
\[(2-4x+4-8x) sin (x-x^2) - (1-2x)^3 cos (x-x^2)\]
\[(6-12x) sin (x-x^2) - (1-2x)^3 cos (x-x^2)\]

is what i get if i kept my head on straight :)

Answer 21

I got a spurious little ^3 thats spose to be a ^2 on the end :)

Answer 22

oo i check again
because the main question is for the power series, so mayb its the D3 that is giving the trouble

Answer 23

ugh, it right ... ^2+^1 = ^3 ...

Answer 24

this just amounts to you doing alot of practice at derivitating stuff :)

Answer 25

ooo D -2cos(x-x^2) is not product rule? == my head is blurrish lol

Answer 26

omg ==

Answer 27

it can be, but you get a 0(cos) which amounts to nothing so we pull the constant and chain the rest

Answer 28

for example

D 2x = D2 x + 2 Dx
= 0 x + 2 1
= 0 + 2
= 2

Answer 29

D 2x = 2 Dx
= 2 1
= 2

Answer 30

ah yes. lol. made silly mistake. currently solving d4

Answer 31

good luck ;)

Answer 32

D4= 12(1-2x)^2 cos(x-x^2) + [(1-2x)^4 -12] sin(x-x^2) ==

Answer 33

http://www.wolframalpha.com/input/?i=4th+derivative+%28sin%28x-x%5E2%29%29

i think i see alot of the parts you got, dbl chk with that to be sure

Answer 34

ooh okay. thx very much for pointing out my mistake ==

Answer 35

erm the cos part, why it is (x-1)x?

Answer 36

its just the way the wolf simplified it in the end ....

Answer 37

owh ic... now i can get the series right... previously i dont know how on earth i can get 2lines worth of pellet for my D3... ==

Answer 38

\[D: (6-12x) sin (x-x^2) - (1-2x)^3 cos (x-x^2)\]
\[D(6-12x) sin (x-x^2)+(6-12x) Dsin (x-x^2) ...\]\[\hspace{10em} - [D(1-2x)^3 cos (x-x^2)+(1-2x)^3 Dcos (x-x^2)]\]

\[-12sin (x-x^2)+(6-12x)(1-x^2) cos(x-x^2) ...\]\[\hspace{10em} - [-6(1-2x)^2 cos (x-x^2)-(1-2x)^4 sin (x-x^2)]\]


\[-12sin (x-x^2)+(6-12x)(1-x^2) cos(x-x^2) ...\]\[\hspace{10em} + 6(1-2x)^2 cos (x-x^2)+(1-2x)^4 sin (x-x^2)\]

and then whatever that simplifies to ...

Answer 39

oh kay thx for your time ^.^ i can go to bed now.

Answer 40

lol, sweet dreams