Question

solve the derivatives (e^3x)/(4x+1) so far i got \[(e ^{3x}(12x-1))/(16x ^{2}+4x+1\]

Answer 1

solve the derivatives? or calculate the derivatives?

Answer 2

\[ e^{3x}(4x+1)^{-1}\]
\[ e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}\]

Answer 3

calculate sorry

Answer 4

\[e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}\]
\[3e^{3x}(4x+1)^{-1} -4e^{3x}(4x+1)^{-2}\]should be the first derivative

Answer 5

it tends to be simpler to do product rule than quotient rule

Answer 6

im really sorry but i dont seem to follow what you did

Answer 7

\[(3e^{3x}(4x+1) -4e^{3x})(4x+1)^{-2}\]
\[\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}\] can be reconstructed

Answer 8

theres an inverse rule \[\frac{1}{a}=a^{-1}\]

Answer 9

i just converted the denominator up into an exponent of -1 to use the product rule on it

Answer 10

ya i multiplied the 3e^xx to the 4x and the 1 and then combined like terms and ultimatly removing a e^3x

Answer 11

if you need more derivatives; its best to keep it in product form

Answer 12

yeah, so far you did fine

Answer 13

how many derivatives you need to take it to? or is the one good enough?

Answer 14

ya its just one i think it says find f'(x)

Answer 15

the ' indicating 1 derivative correct?

Answer 16

then yeah, this is fine; just simplify it as far as you like or leave it as is.

\[\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}\]

as long as you have places to plug in your "x" values to determines slopes thats fine

Answer 17

really?

Answer 18

yes, f' means 1st derivative :)

Answer 19

yeah, prettying it up doesnt change its value one bit

Answer 20

i dont have to simplify it down?

Answer 21

not to use it, no

Answer 22

making something pretty has no consequance on the values it produces

Answer 23

3x + 4x produces the same results for and x value as 7x does right?

Answer 24

you dont think the professor will mark it down

Answer 25

dunno, i aint got your proffessor :)

Answer 26

lol good point well thank you. i must admit i did well for just learning this stuff today

Answer 27

yeah, keep it up :)

Answer 28

even though using this in a physics 100 course seems a bit much

Answer 29

derivatives are use din physics to model certain natural phenomenon and such; its good to know the concepts; but computers and such do the real work these days

Answer 30

ya but teaching this in a course

Answer 31

that is has a co-requisite with calc seems a bit rough, i m in calc im only learning about limists

Answer 32

yeah, limits. something touted about but never really seen again

Answer 33

good because they drive my crazy

Answer 34

epsilons and delta and |x-c| < yada yada ... ugh

Answer 35

its like im just going to be as accurate as i want to be because there is an infinite space between points

Answer 36

well, good luck :)

Answer 37

have a good one and thanks again