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JusaquikieBest 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

KKJBest 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

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

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

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

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

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

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

JusaquikieBest 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

JusaquikieBest 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

KKJBest 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

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

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

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