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

- anonymous

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

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- amistre64

solve the derivatives? or calculate the derivatives?

- amistre64

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

- anonymous

calculate sorry

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## More answers

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

yeah, so far you did fine

- amistre64

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

- anonymous

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

- anonymous

the ' indicating 1 derivative correct?

- amistre64

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

- anonymous

really?

- amistre64

yes, f' means 1st derivative :)

- amistre64

yeah, prettying it up doesnt change its value one bit

- anonymous

i dont have to simplify it down?

- amistre64

not to use it, no

- amistre64

making something pretty has no consequance on the values it produces

- amistre64

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

- anonymous

you dont think the professor will mark it down

- amistre64

dunno, i aint got your proffessor :)

- anonymous

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

- amistre64

yeah, keep it up :)

- anonymous

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

- amistre64

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

- anonymous

ya but teaching this in a course

- anonymous

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

- amistre64

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

- anonymous

good because they drive my crazy

- amistre64

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

- anonymous

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

- amistre64

well, good luck :)

- anonymous

have a good one and thanks again

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