Trisarahtops
  • Trisarahtops
Where is the second derivative of y = 2xe−x equal to 0? 0 1 2 4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Trisarahtops
  • Trisarahtops
@ParthKohli
zepdrix
  • zepdrix
\[\large\rm y=2xe^{-x}\]Do you understand how to find the first derivative? :)
SolomonZelman
  • SolomonZelman
Use the product rule: d/dx (f•g) = f'•g + f•g'

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SolomonZelman
  • SolomonZelman
Do you want an example (with some notes and explanations)?
Trisarahtops
  • Trisarahtops
I got -2e^-x(x-1)
zepdrix
  • zepdrix
For first derivative? Looks good so far :)
SolomonZelman
  • SolomonZelman
the first, that is right...
Trisarahtops
  • Trisarahtops
okay good. now what?
SolomonZelman
  • SolomonZelman
second derivative
SolomonZelman
  • SolomonZelman
differentiate the result (that you obtained for the first derivative). This wll give you the 2nd derivative.
Trisarahtops
  • Trisarahtops
(-2e^-x(x-3) ??
Trisarahtops
  • Trisarahtops
oh no didn't you say it would be 2e^-x(x-2)
SolomonZelman
  • SolomonZelman
yes, checked it, correct
SolomonZelman
  • SolomonZelman
but you should differentiate it yourself, instead of saying; wouldn't you want to do it yourself (tho)?
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle y^{(0)}(x)=kxe^{bx} }\) \(\large\color{#000000 }{ \displaystyle y^{(1)}(x)=ke^{bx}+bkxe^{bx} }\) \(\large\color{#000000 }{ \displaystyle y^{(2)}(x)=ke^{bx}+bke^{bx}+b^2kxe^{bx} }\) \(\large\color{#000000 }{ \displaystyle y^{(2)}(x)=(k+bk)e^{bx}+b^2kxe^{bx} }\)
SolomonZelman
  • SolomonZelman
in general, (with chain rule to the exponent, and the product rule of course)
SolomonZelman
  • SolomonZelman
In any case, you have to set f''(x)=0 and solve for x.
SolomonZelman
  • SolomonZelman
2e^-x(x-2)=0
SolomonZelman
  • SolomonZelman
2e^(-x)\(\ne\)0 \(\forall\)x
SolomonZelman
  • SolomonZelman
So, (x-2)=0 good luck!

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