Jusaquikie
  • Jusaquikie
Find the derivative of the function. y = 5xe^−kx y'(x) =
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Jusaquikie
  • Jusaquikie
\[ y = 5xe^{-kx}\]
Jusaquikie
  • Jusaquikie
i know to use the chain rule and the product rule but i don't know how to seperate this out properly
anonymous
  • anonymous
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

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Jusaquikie
  • Jusaquikie
the answer has ln in it how do e and ln work together?
anonymous
  • anonymous
Where from the In?
anonymous
  • anonymous
If you differentiate y = e^ax, what do you get?
Jusaquikie
  • Jusaquikie
it would be E^ax right?
Jusaquikie
  • Jusaquikie
or would it be ae^ax-1
anonymous
  • anonymous
\[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }\]
Jusaquikie
  • Jusaquikie
or (e^ax)a
anonymous
  • anonymous
it will give you ae^ax
Jusaquikie
  • Jusaquikie
constant letters confuse me more than anything
anonymous
  • anonymous
Consider them as numbers
Jusaquikie
  • Jusaquikie
but moving a to the front of the equason wouldn't i have a -1 on the exponent?
anonymous
  • anonymous
no
Jusaquikie
  • Jusaquikie
so 5xe^-kx is 5x*e^-kx so (5)(e^-kx)+(5x)(-ke^-kx) = 5e^-kx - 5xke^-kx
anonymous
  • anonymous
So \[y=5xe^\left( -kx \right)\] \[\frac{ dy }{ dx } = 5x \frac{ d }{ dx }(e^{-kx}) + e ^{-kx} \frac{ d }{ dx }(5x)\]
anonymous
  • anonymous
\[\frac{ dy }{ dx } = 5x(-ke ^{-kx}) + e ^{-kx}(5)\]
anonymous
  • anonymous
\[\frac{ dy }{ dx } = -5kxe ^{-kx} + 5e ^{-kx}\]
anonymous
  • anonymous
\[\frac{ dy }{ dx } = -5e ^{-kx} (kx - 1)\]
Jusaquikie
  • Jusaquikie
Thanks KKJ

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