Question

Differentiate using the chain rule: (xe^-x)^2

Answer 1

@xapproachesinfinity

Answer 2

well do you know how to use chain rule?

Answer 3

I do, im just having a hard time trying to figure out what the inner and outer function are here

Answer 4

chain rule says for f composition g, F'(x)=f'(g(x))(g'x)

Answer 5

\(\Large \tt\color{blueviolet}{f(x)=(xe^x-x)^2}\)
the inner function is xex-x
the outer let it be x^2

Answer 6

So \(\Large \tt\color{blueviolet}{f'(x)=2(xe^x-x)\times(xe^x-x)'}\)

Answer 7

Im confused as to how you got the first part

Answer 8

well do it like this set f(x)=x^2
g(x)=xe^x-x okay

Answer 9

How did you get g(x)?

Answer 10

now do f'(x)=2x correct?

Answer 11

Since the original problem is\[y=xe ^{-x ^{2}}\]

Answer 12

it is the inner function i just set it equal to g(x) to make this clear

Answer 13

My bad, sorry i typed the original function wrong

Answer 14

oh sorry is though is was \(\Large \tt\color{blueviolet}{xe^x-x}\)

Answer 15

alright let's go back to that function then \(\Large \tt\color{blueviolet}{F(x)=(xe^{-x})
^2}\)

Answer 16

the inner function is \(xe^{-x}\)
the outer ix \(x^2\)

Answer 17

But there are no parentheses around the whole thing

Answer 18

That's why im getting stuck

Answer 19

so what is like \(\Large \tt\color{blueviolet}{xe^{-2x}?}\)

Answer 20

i thought you put parenthesis in your post

Answer 21

\[y=xe ^{-x ^{2}}\]

Answer 22

oh i see! you first need to do product rule

Answer 23

Oh ok that makes sense

Answer 24

\(\Large \tt\color{blueviolet}{(xe^{-x^2})'=x'e^{-x^2}+x(e^{-x^2})'}\)

Answer 25

x'=1
you just need to do chain rule to the last term. yes?

Answer 26

Right, which would be \[2e^{-x}(-e^{-x})\]?
Then it would be \[-2e^{-2x}\] ?

Answer 27

hmm not really set f(x)=e^x
ang g(x)=-x^2 and see

Answer 28

it would be \(\Large \tt\color{blueviolet}{-2xe^{-x^2}}\) agree?

Answer 29

Oh ok yep that makes more sense

Answer 30

now you just need to clean that thing and factor exp(-x^2) and that's all

Answer 31

remember you sill have x attached to (e^{-x^2})

Answer 32

you get it?