Question

if y=u+2e^u and u=1+lnx, find dy/dx when x=(1/e), please help i got as far as plugging everything in and getting y=1+ln(1/e)+2e^(1+ln(1/e))

Answer 1

use chain rule
\[\frac{ dy }{dx }=\frac{dy}{du}\times \frac{du}{dx}\]

Answer 2

im not sure how to do that in this case

Answer 3

oh wait, please hold on one moment

Answer 4

ok

Answer 5

like I showed in the formula
differentiate y with respect to u
differentiate u with respect to x

multiply the two, you get dy/dx

Answer 6

would you recommend differenciating every value first?

Answer 7

for example if we know x=1/e then we use quotient rule

Answer 8

noo, x is a variable,
you can substitute x=1/e only after you have found dy/dx

Answer 9

remember 'e' is a constant

Answer 10

not another variable

Answer 11

so we begin by finding the derivative of y=u+2e^u?

Answer 12

yes, with respect to 'u'

Answer 13

that means we plug in u already

Answer 14

no, consider 'u' as a variable

Answer 15

we are not plugging in anything

Answer 16

ok, so when i differentiated y=u+=2e^u i got
u+2(lne)(e^u)(u')

Answer 17

and (lne)=1

Answer 18

y=u + \(2e^u\)

\[\frac{dy}{du}=1 + 2e^u\]

Answer 19

so we don't have to find the derivative of 2e^u?

Answer 20

according to the rules of differentiation\[\frac{ d(e^x) }{ dx }=e^x\]

Answer 21

ok so what would the next step be?

Answer 22

do we plug in u now?

Answer 23

no,
now
u= 1 +ln x

find du/dx

Answer 24

1/x

Answer 25

yes

now multiply dy/du and du/dx

Answer 26

you mean (1+2e^u)*(1/x)?

Answer 27

yes,

now that is dy/dx

Answer 28

\[\frac{dy}{dx}=(1+2e^u)\times \frac{1}{x}\]

Answer 29

would it be (1/x)(2e^u/x) and now that they are over the same denomentator then 2e^u

Answer 30

i dont know what you mean...
but have you followed till
dy/dx ?

Answer 31

yes (1 * 2e^u) * 1/x

Answer 32

now i can distribute

Answer 33

right?

Answer 34

now, you need to plug these in
\[x=1/e \\ u=1+\ln(1/e)\]

you can simplify u a bit further before you plug in

Answer 35

i got \[(1+2e^(1+\ln(1/x)))(1/(1/e))\]

Answer 36

You have to put curly braces around the stuff you want in exponents :)
e^{1+ln(1/x)}
Sall good though, LaTeX takes a little getting used to ^^

Answer 37

oh thank you

Answer 38

i dont follow what you did,
how do you still have an 'x' in there?

Answer 39

\[(1+2e^{1+\ln(1/(1/e))})(1/(1/e))\]

Answer 40

have a look at this

\[u=1 + \ln(1/e) \\ u=1+\ln(e^{-1})\\u=1-\ln (e)\\u=1-1\\u=0\]

Answer 41

oh goodness i didn't think of that

Answer 42

\[\frac{dy}{dx}=(1+2e^u)\times \frac{1}{x} \\ plug ~ In\\ x=1/e ~and ~u=0\]

Answer 43

2e

Answer 44

check again...
i get 3e

Answer 45

oh i see how becasue 2e^0 is 2(e^0) and so it is 2(1)

Answer 46

yep :)

Answer 47

thank you so much

Answer 48

sure :)

Answer 49

sorry it took so long

Answer 50

not at all :p

Answer 51

@zepdrix anything to add?

Answer 52

Nahhh c:
gj guysss

Answer 53

cool :P
cuz u were making me a bit insecure xD

Answer 54

no way you were doing a great job