Question

How is F'' of (1-x)e^-x
Equal to (x-2)e^-x ?

Answer 1

if F(x) = (1-x)e^-x
then the double differential is that,

using F' = u(dv/dx) + v(du/dx)

where u= (1-x) v=e^-x


then do the same method again to work out the F''

Answer 2

Ok product rule then, well its still not working!

Answer 3

I get (0)(e^-x)+(1-x)(e^-x)

Answer 4

u=(1-x)
du/dx=-1

v=e^-x
dv/dx=-e^-x

Answer 5

omg of course :) thanks!

Answer 6

but wait how come the - sign can be used 2x?

Answer 7

i dont understand your q

Answer 8

I have (1-x^1) I bring the 1 down in front of the x and its (1-1)?

Answer 9

how does it become (1-(-1)?

Answer 10

the differential of a constant is zero , 0

Answer 11

ok so its -1 then not -2? where does the -2 come from in (x-2)?

Answer 12

you have to differentiate again

F' is dy/dx
F'' is d^2y/dx^2

F''' is d^3y/dx^3
;;;
;;;
F^n is d^ny/dx^n

Answer 13

yes I am diferentiating again
F'' to me is I get (0)(e^-x)+(1-x)(e^-x) using the product rule?

Answer 14

ok what about (0)(-1)e^-x*(1-x)-e^-x

Answer 15

Can you please tell me what to do I can't spend 30 min on this one stupid derivative!

Answer 16

okay

Answer 17

oh the original question is xe^-x
sad part is this is an example in the book I don't even need to do it I just want to understand it

Answer 18

let
u=1-x
u'=-1

v=e^-x
v'=-e^-x

then uv'+vu'=

[1-x](-)e^-x + (-1)e^-x

so

e^-x[(x-1)-1]

e^-x [x-2]


then again do the same thing


let
u=e^-x
u'=-e^-x

v=x-2
v'=1

uv'+vu'

=

e^-x[1] + [x-2](-1)e^-x
=

e^-x *[ 1+2-x]

=
e^-x *[3-x]


mmm, can you spot where i went wrong?

Answer 19

How does u=1-x equal u'=-1
I get 0 when I do that

Answer 20

oh crap nevermind! UGH I hate this sometimes. the constant is factored out

Answer 21

the differential of -x is -1*x
so we are left with -1

if you choose u=-1
v=x

then v'=1
u'=0

uv'+vu' = (-1)*1 + x*(0) = -1

Answer 22

Last question, for f' of xe^-x how does the x and the -1 get grouped like (1-x)
When I get f' I get (-1)(x)e^-x

Answer 23

u=x
u'=1

v=e^-x
v'=-e^-x

uv'+vu'=

x(-1)e^-x + (1)*e^-x

=

e^-x* [-x+1]

Answer 24

Thanks Rezz you are the man or woman!

Answer 25

yw!!
im a man btw

PS, if you get stuck with differentiation always use product rule, it also works with using in fractions, so for example you dont need to use the quotient rule just product rule, im in 2nd uni doing maths and i stil dont use quotient rule!!!

^
|


was for your own information!

Answer 26

I didnt realize that when using u'(g)+u(g') you could use the chain rule on "g" first and THEN you get g'. I was splitting g' up into -x and then e^-x as two seperate things, if this makes any sense

Answer 27

you can use quotient rule for the first question where e^-x = 1/e^x

Answer 28

no idea!

Answer 29

haha thats ok