Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

right!

OpenStudy (anonymous):

$\frac{e^{2x}\times 2-2x\times 2e^{2x}}{4x^2}$

OpenStudy (anonymous):

got it

OpenStudy (anonymous):

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

OpenStudy (anonymous):

oh no, you did not do that sorry. hmmmmm

OpenStudy (anonymous):

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

OpenStudy (anonymous):

hope that is a choice!

OpenStudy (anonymous):

B, it is B

OpenStudy (anonymous):

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!!!!