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

Mathematics
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\[ y = 5xe^{-kx}\]
i know to use the chain rule and the product rule but i don't know how to seperate this out properly
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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Other answers:

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

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