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anonymous
 4 years ago
solve the derivatives (e^3x)/(4x+1) so far i got \[(e ^{3x}(12x1))/(16x ^{2}+4x+1\]
anonymous
 4 years ago
solve the derivatives (e^3x)/(4x+1) so far i got \[(e ^{3x}(12x1))/(16x ^{2}+4x+1\]

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1solve the derivatives? or calculate the derivatives?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[ e^{3x}(4x+1)^{1}\] \[ e'^{3x}(4x+1)^{1}+ e^{3x}(4x+1)'^{1}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[e'^{3x}(4x+1)^{1}+ e^{3x}(4x+1)'^{1}\] \[3e^{3x}(4x+1)^{1} 4e^{3x}(4x+1)^{2}\]should be the first derivative

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it tends to be simpler to do product rule than quotient rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im really sorry but i dont seem to follow what you did

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[(3e^{3x}(4x+1) 4e^{3x})(4x+1)^{2}\] \[\frac{3e^{3x}(4x+1) 4e^{3x}}{(4x+1)^{2}}\] can be reconstructed

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1theres an inverse rule \[\frac{1}{a}=a^{1}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i just converted the denominator up into an exponent of 1 to use the product rule on it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya i multiplied the 3e^xx to the 4x and the 1 and then combined like terms and ultimatly removing a e^3x

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if you need more derivatives; its best to keep it in product form

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, so far you did fine

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1how many derivatives you need to take it to? or is the one good enough?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya its just one i think it says find f'(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the ' indicating 1 derivative correct?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1then 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

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, f' means 1st derivative :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, prettying it up doesnt change its value one bit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont have to simplify it down?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1making something pretty has no consequance on the values it produces

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.13x + 4x produces the same results for and x value as 7x does right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you dont think the professor will mark it down

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1dunno, i aint got your proffessor :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol good point well thank you. i must admit i did well for just learning this stuff today

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0even though using this in a physics 100 course seems a bit much

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1derivatives 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
 4 years ago
Best ResponseYou've already chosen the best response.0ya but teaching this in a course

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is has a corequisite with calc seems a bit rough, i m in calc im only learning about limists

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, limits. something touted about but never really seen again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0good because they drive my crazy

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1epsilons and delta and xc < yada yada ... ugh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its like im just going to be as accurate as i want to be because there is an infinite space between points

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have a good one and thanks again
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