find the derivative of f(x)=e^2x/2x? i keep getting stuck.... first i get f'(x)= e^(2x) (2)- (2x) (2e^(2x)) all of this is /(2x)^2 then i simplyfiy it and get 2e^(2x)-4xe^(2x) and all of it /2x^2 can i simplfy it more or am i doing something wrong cause i'm not getting any of the answer choices that are given like A)1 B)(e^(2x)(1-2x))/2x^2 C) e^2x D) i know this is wrong E) (e^(2x) (2x-1))/2x^2

quotent rule \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\]

\[f(x)=e^{2x},f'(x)=2e^{2x}\] \[g(x)=2x,g'(x)=2\] and put this together right?

right!

\[\frac{e^{2x}\times 2-2x\times 2e^{2x}}{4x^2}\]

got it

it is easy to forget the chain rule when you just have a constant times the variable. like in \[\sin(3x)\] or \[e^{-x}\]

oh no, you did not do that sorry. hmmmmm

k now we simplify \[\frac{2e^{2x}-4xe^{2x}}{4x^2}\] \[\frac{2e^{2x}(1-2x)}{4x^2}\] \[\frac{e^{2x}(1-2x)}{2x^2}\]

hope that is a choice!

B, it is B

yup it is thanks i see what i did wrong... i messed up the bottom of the equation... thanks you were a lot of help!!!!

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