Is this correct

  • #1
y=-1/2 x cosx



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'
 

Answers and Replies

  • #2
169
0
I am an amateur myself but I recall (hopefully) that cos(x) differentiates to -sin(x)

EDIT: I didn't see the negative sorry
 
  • #3
statdad
Homework Helper
1,495
35
If your first derivative is

[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
Will the second derivative be
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
Well I'm doing differential equations and I have to match it with Y" + y= sin x
 
  • #6
169
0
y=-1/2 x cosx



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

I didn't get the ^2x part.
 
  • #7
y''= (1/2)xcosx + sinx

I didn't get the ^2x part.
How'd you get that second derivative. I did it wrong how'd u do it, you did it right
 
  • #8
When you add Y"+y=sin x




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

which becomes sin x
 

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