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anonymous

  • 4 years ago

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

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  1. amistre64
    • 4 years ago
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    solve the derivatives? or calculate the derivatives?

  2. amistre64
    • 4 years ago
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    \[ e^{3x}(4x+1)^{-1}\] \[ e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}\]

  3. anonymous
    • 4 years ago
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    calculate sorry

  4. amistre64
    • 4 years ago
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    \[e'^{3x}(4x+1)^{-1}+ e^{3x}(4x+1)'^{-1}\] \[3e^{3x}(4x+1)^{-1} -4e^{3x}(4x+1)^{-2}\]should be the first derivative

  5. amistre64
    • 4 years ago
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    it tends to be simpler to do product rule than quotient rule

  6. anonymous
    • 4 years ago
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    im really sorry but i dont seem to follow what you did

  7. amistre64
    • 4 years ago
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    \[(3e^{3x}(4x+1) -4e^{3x})(4x+1)^{-2}\] \[\frac{3e^{3x}(4x+1) -4e^{3x}}{(4x+1)^{2}}\] can be reconstructed

  8. amistre64
    • 4 years ago
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    theres an inverse rule \[\frac{1}{a}=a^{-1}\]

  9. amistre64
    • 4 years ago
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    i just converted the denominator up into an exponent of -1 to use the product rule on it

  10. anonymous
    • 4 years ago
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    ya i multiplied the 3e^xx to the 4x and the 1 and then combined like terms and ultimatly removing a e^3x

  11. amistre64
    • 4 years ago
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    if you need more derivatives; its best to keep it in product form

  12. amistre64
    • 4 years ago
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    yeah, so far you did fine

  13. amistre64
    • 4 years ago
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    how many derivatives you need to take it to? or is the one good enough?

  14. anonymous
    • 4 years ago
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    ya its just one i think it says find f'(x)

  15. anonymous
    • 4 years ago
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    the ' indicating 1 derivative correct?

  16. amistre64
    • 4 years ago
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    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

  17. anonymous
    • 4 years ago
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    really?

  18. amistre64
    • 4 years ago
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    yes, f' means 1st derivative :)

  19. amistre64
    • 4 years ago
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    yeah, prettying it up doesnt change its value one bit

  20. anonymous
    • 4 years ago
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    i dont have to simplify it down?

  21. amistre64
    • 4 years ago
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    not to use it, no

  22. amistre64
    • 4 years ago
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    making something pretty has no consequance on the values it produces

  23. amistre64
    • 4 years ago
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    3x + 4x produces the same results for and x value as 7x does right?

  24. anonymous
    • 4 years ago
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    you dont think the professor will mark it down

  25. amistre64
    • 4 years ago
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    dunno, i aint got your proffessor :)

  26. anonymous
    • 4 years ago
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    lol good point well thank you. i must admit i did well for just learning this stuff today

  27. amistre64
    • 4 years ago
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    yeah, keep it up :)

  28. anonymous
    • 4 years ago
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    even though using this in a physics 100 course seems a bit much

  29. amistre64
    • 4 years ago
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    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

  30. anonymous
    • 4 years ago
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    ya but teaching this in a course

  31. anonymous
    • 4 years ago
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    that is has a co-requisite with calc seems a bit rough, i m in calc im only learning about limists

  32. amistre64
    • 4 years ago
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    yeah, limits. something touted about but never really seen again

  33. anonymous
    • 4 years ago
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    good because they drive my crazy

  34. amistre64
    • 4 years ago
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    epsilons and delta and |x-c| < yada yada ... ugh

  35. anonymous
    • 4 years ago
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    its like im just going to be as accurate as i want to be because there is an infinite space between points

  36. amistre64
    • 4 years ago
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    well, good luck :)

  37. anonymous
    • 4 years ago
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    have a good one and thanks again

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