anonymous
  • anonymous
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))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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baru
  • baru
use chain rule \[\frac{ dy }{dx }=\frac{dy}{du}\times \frac{du}{dx}\]
anonymous
  • anonymous
im not sure how to do that in this case
anonymous
  • anonymous
oh wait, please hold on one moment

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baru
  • baru
ok
baru
  • baru
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
anonymous
  • anonymous
would you recommend differenciating every value first?
anonymous
  • anonymous
for example if we know x=1/e then we use quotient rule
baru
  • baru
noo, x is a variable, you can substitute x=1/e only after you have found dy/dx
baru
  • baru
remember 'e' is a constant
baru
  • baru
not another variable
anonymous
  • anonymous
so we begin by finding the derivative of y=u+2e^u?
baru
  • baru
yes, with respect to 'u'
anonymous
  • anonymous
that means we plug in u already
baru
  • baru
no, consider 'u' as a variable
baru
  • baru
we are not plugging in anything
anonymous
  • anonymous
ok, so when i differentiated y=u+=2e^u i got u+2(lne)(e^u)(u')
anonymous
  • anonymous
and (lne)=1
baru
  • baru
y=u + \(2e^u\) \[\frac{dy}{du}=1 + 2e^u\]
anonymous
  • anonymous
so we don't have to find the derivative of 2e^u?
baru
  • baru
according to the rules of differentiation\[\frac{ d(e^x) }{ dx }=e^x\]
anonymous
  • anonymous
ok so what would the next step be?
anonymous
  • anonymous
do we plug in u now?
baru
  • baru
no, now u= 1 +ln x find du/dx
anonymous
  • anonymous
1/x
baru
  • baru
yes now multiply dy/du and du/dx
anonymous
  • anonymous
you mean (1+2e^u)*(1/x)?
baru
  • baru
yes, now that is dy/dx
baru
  • baru
\[\frac{dy}{dx}=(1+2e^u)\times \frac{1}{x}\]
anonymous
  • anonymous
would it be (1/x)(2e^u/x) and now that they are over the same denomentator then 2e^u
baru
  • baru
i dont know what you mean... but have you followed till dy/dx ?
anonymous
  • anonymous
yes (1 * 2e^u) * 1/x
anonymous
  • anonymous
now i can distribute
anonymous
  • anonymous
right?
baru
  • baru
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
anonymous
  • anonymous
i got \[(1+2e^(1+\ln(1/x)))(1/(1/e))\]
zepdrix
  • zepdrix
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 ^^
anonymous
  • anonymous
oh thank you
baru
  • baru
i dont follow what you did, how do you still have an 'x' in there?
anonymous
  • anonymous
\[(1+2e^{1+\ln(1/(1/e))})(1/(1/e))\]
baru
  • baru
have a look at this \[u=1 + \ln(1/e) \\ u=1+\ln(e^{-1})\\u=1-\ln (e)\\u=1-1\\u=0\]
anonymous
  • anonymous
oh goodness i didn't think of that
baru
  • baru
\[\frac{dy}{dx}=(1+2e^u)\times \frac{1}{x} \\ plug ~ In\\ x=1/e ~and ~u=0\]
anonymous
  • anonymous
2e
baru
  • baru
check again... i get 3e
anonymous
  • anonymous
oh i see how becasue 2e^0 is 2(e^0) and so it is 2(1)
baru
  • baru
yep :)
anonymous
  • anonymous
thank you so much
baru
  • baru
sure :)
anonymous
  • anonymous
sorry it took so long
baru
  • baru
not at all :p
baru
  • baru
@zepdrix anything to add?
zepdrix
  • zepdrix
Nahhh c: gj guysss
baru
  • baru
cool :P cuz u were making me a bit insecure xD
anonymous
  • anonymous
no way you were doing a great job

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