Exploring y", y" with the Product Rule

[afcwestwarrior]

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'

 

[Unassuming]

I am an amateur myself but I recall (hopefully) that cos(x) differentiates to -sin(x)

EDIT: I didn't see the negative sorry

 

[statdad]

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.

 

[afcwestwarrior]

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

 

[afcwestwarrior]

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

 

[Unassuming]

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.

 

[afcwestwarrior]

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

 

[afcwestwarrior]

When you add Y"+y=sin x




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

which becomes sin x