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Jusaquikie

Find the derivative of the function. y = 5xe^−kx y'(x) =

  • one year ago
  • one year ago

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  1. Jusaquikie
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    \[ y = 5xe^{-kx}\]

    • one year ago
  2. Jusaquikie
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    i know to use the chain rule and the product rule but i don't know how to seperate this out properly

    • one year ago
  3. KKJ
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    y is made up of a product function. Thus, 5x and e^(-kx) Hence, we are to use the product rule with the above separations

    • one year ago
  4. Jusaquikie
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    the answer has ln in it how do e and ln work together?

    • one year ago
  5. KKJ
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    Where from the In?

    • one year ago
  6. KKJ
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    If you differentiate y = e^ax, what do you get?

    • one year ago
  7. Jusaquikie
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    it would be E^ax right?

    • one year ago
  8. Jusaquikie
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    or would it be ae^ax-1

    • one year ago
  9. KKJ
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    \[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }\]

    • one year ago
  10. Jusaquikie
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    or (e^ax)a

    • one year ago
  11. KKJ
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    it will give you ae^ax

    • one year ago
  12. Jusaquikie
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    constant letters confuse me more than anything

    • one year ago
  13. KKJ
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    Consider them as numbers

    • one year ago
  14. Jusaquikie
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    but moving a to the front of the equason wouldn't i have a -1 on the exponent?

    • one year ago
  15. KKJ
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    no

    • one year ago
  16. Jusaquikie
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    so 5xe^-kx is 5x*e^-kx so (5)(e^-kx)+(5x)(-ke^-kx) = 5e^-kx - 5xke^-kx

    • one year ago
  17. KKJ
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    So \[y=5xe^\left( -kx \right)\] \[\frac{ dy }{ dx } = 5x \frac{ d }{ dx }(e^{-kx}) + e ^{-kx} \frac{ d }{ dx }(5x)\]

    • one year ago
  18. KKJ
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    \[\frac{ dy }{ dx } = 5x(-ke ^{-kx}) + e ^{-kx}(5)\]

    • one year ago
  19. KKJ
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    \[\frac{ dy }{ dx } = -5kxe ^{-kx} + 5e ^{-kx}\]

    • one year ago
  20. KKJ
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    \[\frac{ dy }{ dx } = -5e ^{-kx} (kx - 1)\]

    • one year ago
  21. Jusaquikie
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    Thanks KKJ

    • one year ago
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