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



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

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

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

Jusaquikie
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the answer has ln in it how do e and ln work together?

KKJ
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Where from the In?

KKJ
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If you differentiate y = e^ax, what do you get?

Jusaquikie
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it would be E^ax right?

Jusaquikie
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or would it be ae^ax1

KKJ
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\[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }\]

Jusaquikie
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or (e^ax)a

KKJ
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it will give you ae^ax

Jusaquikie
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constant letters confuse me more than anything

KKJ
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Consider them as numbers

Jusaquikie
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but moving a to the front of the equason wouldn't i have a 1 on the exponent?

KKJ
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no

Jusaquikie
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so 5xe^kx is 5x*e^kx so
(5)(e^kx)+(5x)(ke^kx) =
5e^kx  5xke^kx

KKJ
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So \[y=5xe^\left( kx \right)\]
\[\frac{ dy }{ dx } = 5x \frac{ d }{ dx }(e^{kx}) + e ^{kx} \frac{ d }{ dx }(5x)\]

KKJ
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\[\frac{ dy }{ dx } = 5x(ke ^{kx}) + e ^{kx}(5)\]

KKJ
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\[\frac{ dy }{ dx } = 5kxe ^{kx} + 5e ^{kx}\]

KKJ
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\[\frac{ dy }{ dx } = 5e ^{kx} (kx  1)\]

Jusaquikie
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Thanks KKJ