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Jusaquikie

  • 2 years ago

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

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  1. Jusaquikie
    • 2 years ago
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    \[ y = 5xe^{-kx}\]

  2. Jusaquikie
    • 2 years ago
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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

  3. KKJ
    • 2 years ago
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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

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

  5. KKJ
    • 2 years ago
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    Where from the In?

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

  7. Jusaquikie
    • 2 years ago
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    it would be E^ax right?

  8. Jusaquikie
    • 2 years ago
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    or would it be ae^ax-1

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

  10. Jusaquikie
    • 2 years ago
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    or (e^ax)a

  11. KKJ
    • 2 years ago
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    it will give you ae^ax

  12. Jusaquikie
    • 2 years ago
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    constant letters confuse me more than anything

  13. KKJ
    • 2 years ago
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    Consider them as numbers

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

  15. KKJ
    • 2 years ago
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    no

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

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

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

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

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

  21. Jusaquikie
    • 2 years ago
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    Thanks KKJ

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