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Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0i know to use the chain rule and the product rule but i don't know how to seperate this out properly

KKJ
 2 years ago
Best ResponseYou've already chosen the best response.1y 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
 2 years ago
Best ResponseYou've already chosen the best response.0the answer has ln in it how do e and ln work together?

KKJ
 2 years ago
Best ResponseYou've already chosen the best response.1If you differentiate y = e^ax, what do you get?

Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0it would be E^ax right?

Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0or would it be ae^ax1

KKJ
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }\]

Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0constant letters confuse me more than anything

Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0but moving a to the front of the equason wouldn't i have a 1 on the exponent?

Jusaquikie
 2 years ago
Best ResponseYou've already chosen the best response.0so 5xe^kx is 5x*e^kx so (5)(e^kx)+(5x)(ke^kx) = 5e^kx  5xke^kx

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

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

KKJ
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = 5kxe ^{kx} + 5e ^{kx}\]

KKJ
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = 5e ^{kx} (kx  1)\]
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