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Jusaquikie
Group Title
Find the derivative of the function.
y = 5xe^−kx
y'(x) =
 one year ago
 one year ago
Jusaquikie Group Title
Find the derivative of the function. y = 5xe^−kx y'(x) =
 one year ago
 one year ago

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Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
\[ y = 5xe^{kx}\]
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
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

KKJ Group TitleBest ResponseYou've already chosen the best response.1
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

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
the answer has ln in it how do e and ln work together?
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
Where from the In?
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
If you differentiate y = e^ax, what do you get?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
it would be E^ax right?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
or would it be ae^ax1
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }\]
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
or (e^ax)a
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
it will give you ae^ax
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
constant letters confuse me more than anything
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
Consider them as numbers
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
but moving a to the front of the equason wouldn't i have a 1 on the exponent?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
so 5xe^kx is 5x*e^kx so (5)(e^kx)+(5x)(ke^kx) = 5e^kx  5xke^kx
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
So \[y=5xe^\left( kx \right)\] \[\frac{ dy }{ dx } = 5x \frac{ d }{ dx }(e^{kx}) + e ^{kx} \frac{ d }{ dx }(5x)\]
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx } = 5x(ke ^{kx}) + e ^{kx}(5)\]
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx } = 5kxe ^{kx} + 5e ^{kx}\]
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx } = 5e ^{kx} (kx  1)\]
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
Thanks KKJ
 one year ago
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