Question

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

Answer 1

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

Answer 2

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

Answer 3

right!

Answer 4

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

Answer 5

got it

Answer 6

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

Answer 7

oh no, you did not do that sorry. hmmmmm

Answer 8

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

Answer 9

hope that is a choice!

Answer 10

B, it is B

Answer 11

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!!!!