Question

How do you solve for the derivative of (e^2x)/2x

Answer 1

It is of the form f(x)/g(x)

Use quotient rule.

Answer 2

Ive set it up as \[\frac{ 2x(2xe ^{2x})- e ^{2x}(2) }{ 2x^2 }\]

Answer 3

i have tried reducing it but its not any of the choices on my homework. they seem to factor a 2 or 2x out but i cant get the inside.

Answer 4

downstair shoudl be 4x^2?

Answer 5

I am rusty on calc though

Answer 6

okay, i will try that

Answer 7

hmm... I cant seem too get an any of the answers using \[4x^2\]either...

Answer 8

what are the choices?

Answer 9

a. \[\frac{ 2(2xe-2) }{ x }\]

Answer 10

b. \[\frac{ 2e(2x-2) }{ 2x }\]

Answer 11

what? e on its own, not e^x or something?

Answer 12

ohh sorry i forgot a the ^2, it should be 2e^2

Answer 13

c. \[\frac{ 4x(e^2-2) }{ 2x^2 }\]

Answer 14

I'll go ask Wolfram

Answer 15

Thanks

Answer 16

你好
Nei Hou everyone XD

@caozeyuan
Giving up so soon? :>

Answer 17

d. \[\frac{ 2x(e^2-2)}{ 2x^2 }\]

Answer 18

\[\frac{ e^{2x}(2x-1) }{ 2x ^{2} }\]

Answer 19

As it's already been mentioned, we use the quotient rule: \[\dfrac{d}{dx}(\dfrac{f(x)}{g(x)}) = \dfrac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\]Where \[f(x)=e^{2x}\]\[g(x)=2x\]So\[f'(x)=2e^{2x}\]\[g'(x)=2\]Giving us\[\dfrac{2e^{2x}\times2x-e^{2x}\times2}{4x^2} = \dfrac{4xe^{2x}-2e^{2x}}{4x^2} = \dfrac{(2x-1)e^{2x}}{2x^2}\]

Answer 20

That last answer you posted is right @caozeyuan

Answer 21

Its (2x)^2 not 2x^2

Answer 22

A typo in the choices, then?

Answer 23

Okay i see that i think the e^2x confused me a bit in my math. Thank you :) @tom982

Answer 24

Yep, none of the choices make sense to me

Answer 25

No problem Johan :)

Answer 26

It could be that my teacher was excited for the thanksgiving break. Took a test on our last day some of the answers where repeated... so not a very reliable way to check my work.

Answer 27

Thanks everyone, i will trudge through the rest of my homework now C:

Answer 28

Hope it goes well. Drop me a message if you want me to have a look at something.