- #1

- 457

- 0

y'=1/2 x sinx + cos x -1/2

y"= 1/2 x cosx - sin x 1/2 + cos x + -1/2 - sin x

y"= 1/2 x cos^2x

I used the product rule (fg)'=fg'+gf'

- Thread starter afcwestwarrior
- Start date

- #1

- 457

- 0

y'=1/2 x sinx + cos x -1/2

y"= 1/2 x cosx - sin x 1/2 + cos x + -1/2 - sin x

y"= 1/2 x cos^2x

I used the product rule (fg)'=fg'+gf'

- #2

- 169

- 0

EDIT: I didn't see the negative sorry

- #3

statdad

Homework Helper

- 1,495

- 35

[tex]

y' = \frac 1 2 x \sin x - \frac 1 2 \cos x

[/tex]

that is correct.

I do not believe your second derivative is correct.

- #4

- 457

- 0

y"=1/2 x cos x + sin x * 1/2 - 1/2 - sin x

y"=1/2 x cos x + 1/2 sin x + 1/2 sin x

- #5

- 457

- 0

Well I'm doing differential equations and I have to match it with Y" + y= sin x

- #6

- 169

- 0

y''= (1/2)xcosx + sinx

y'=1/2 x sinx + cos x -1/2

y"= 1/2 x cosx - sin x 1/2 + cos x + -1/2 - sin x

y"= 1/2 x cos^2x

I used the product rule (fg)'=fg'+gf'

I didn't get the ^2x part.

- #7

- 457

- 0

How'd you get that second derivative. I did it wrong how'd u do it, you did it righty''= (1/2)xcosx + sinx

I didn't get the ^2x part.

- #8

- 457

- 0

When you add Y"+y=sin x

(1/2)xcosx + sinx + -1/2 x cosx

which becomes sin x

(1/2)xcosx + sinx + -1/2 x cosx

which becomes sin x

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