anonymous
  • anonymous
solve the derivatives (e^3x)/(4x+1) so far i got \[(e ^{3x}(12x-1))/(16x ^{2}+4x+1\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
solve the derivatives? or calculate the derivatives?
amistre64
  • amistre64
\[ e^{3x}(4x+1)^{-1}\] \[ e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}\]
anonymous
  • anonymous
calculate sorry

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amistre64
  • 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
  • amistre64
it tends to be simpler to do product rule than quotient rule
anonymous
  • anonymous
im really sorry but i dont seem to follow what you did
amistre64
  • amistre64
\[(3e^{3x}(4x+1) -4e^{3x})(4x+1)^{-2}\] \[\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}\] can be reconstructed
amistre64
  • amistre64
theres an inverse rule \[\frac{1}{a}=a^{-1}\]
amistre64
  • amistre64
i just converted the denominator up into an exponent of -1 to use the product rule on it
anonymous
  • 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
  • amistre64
if you need more derivatives; its best to keep it in product form
amistre64
  • amistre64
yeah, so far you did fine
amistre64
  • amistre64
how many derivatives you need to take it to? or is the one good enough?
anonymous
  • anonymous
ya its just one i think it says find f'(x)
anonymous
  • anonymous
the ' indicating 1 derivative correct?
amistre64
  • 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
  • anonymous
really?
amistre64
  • amistre64
yes, f' means 1st derivative :)
amistre64
  • amistre64
yeah, prettying it up doesnt change its value one bit
anonymous
  • anonymous
i dont have to simplify it down?
amistre64
  • amistre64
not to use it, no
amistre64
  • amistre64
making something pretty has no consequance on the values it produces
amistre64
  • amistre64
3x + 4x produces the same results for and x value as 7x does right?
anonymous
  • anonymous
you dont think the professor will mark it down
amistre64
  • amistre64
dunno, i aint got your proffessor :)
anonymous
  • anonymous
lol good point well thank you. i must admit i did well for just learning this stuff today
amistre64
  • amistre64
yeah, keep it up :)
anonymous
  • anonymous
even though using this in a physics 100 course seems a bit much
amistre64
  • 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
  • anonymous
ya but teaching this in a course
anonymous
  • anonymous
that is has a co-requisite with calc seems a bit rough, i m in calc im only learning about limists
amistre64
  • amistre64
yeah, limits. something touted about but never really seen again
anonymous
  • anonymous
good because they drive my crazy
amistre64
  • amistre64
epsilons and delta and |x-c| < yada yada ... ugh
anonymous
  • anonymous
its like im just going to be as accurate as i want to be because there is an infinite space between points
amistre64
  • amistre64
well, good luck :)
anonymous
  • anonymous
have a good one and thanks again

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