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Simplifying the derivative after using the Product Rule. Help?

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    f(x) = (x^3+6)^4 * (x^3+4)^6


    2. Relevant equations
    f'(x) = (x^3+6)^4*18x^2(x^3+4)^5 + (x^3+4)^6*12x^2(x^3+6)^3


    3. The attempt at a solution
    Obviously this expression can be simplified, but I haven't the slightest clue on how to do this.

    I would really appreciate the help!
     
  2. jcsd
  3. Dec 6, 2011 #2

    Mark44

    Staff: Mentor

    Your derivative is made up of two terms:
    18x2(x3+4)5(x3+6)4
    and
    12x2(x3+4)6(x3+6)3

    Obviously, both terms have a factor of 6x2. Both terms also have factors of (x3 + 4) and (x3 + 6).

    For each of these factors, bring out the smallest power that is present in both terms.

    You should end up with 6x2(x3 + 4)m(x3 + 6)n(something + something else)
     
  4. Dec 6, 2011 #3
    I'm beginning to understand. If you don't mind, I'm going to show you what I did step by step... Hopefully I do it right.

    6x2(x3+4)5(x3+6)3* 3(x3+6) + 12(x3+4) =

    6x2(x3+4)5(x3+6)3* (3x3+12+12x3+48) =

    6x2(x3+4)5(x3+6)3*(15x3+60)

    Did I do the simplifying right?
     
  5. Dec 6, 2011 #4

    Mark44

    Staff: Mentor

    You have a mistake in the line above, and that affects everything below here.
    6x2(x3+4)5(x3+6)3* [3(x3+6) + [STRIKE]12[/STRIKE]2(x3+4)]

    Can you take it from here? There's also another mistake below - 3*6 = 18, not 12.
     
  6. Dec 6, 2011 #5

    epenguin

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    Gold Member

    When I was a young student I used to think there was virtue in long formulae like those and if I could handle them it showed I was smart. If anything it showed I wasn't. Took me quite some time to cotton on to the where where where of mathematicians.

    I.e. saying instead of your formulae f(x) = 18x2(x3+4)5(x3+6)4 you can say differentiate 18x2u5v4 where u = (x3+4) and v = (x3+6).

    Then
    f'(x) = 18( 2xu5v4 + 5x2u4u'v4 + 4x2u5v3v') .

    You can already see you can take out a factor xu4v3, and further down there may be more simplifications. That way you see what you are doing, the essentials of it. (Of course you do have to remember that the u, v are functions of x so you have to use the chain rule and write not 5u4 but 5u4u' .) Then at the end of the calculation you'll often enough convert back to original variables.

    There is no virtue in unnecessarily complicated expressions (and while I am at it, unnecessarily general ones either - you will soon come up across the expression 'without loss of generality'.)
     
  7. Dec 6, 2011 #6
    I always make basic errors.

    Alright. So with those fixed, this is what my simplified derivative came out to be:

    6x2(x3+4)5(x3+6)3(5x3+26)

    Better?

    Also, THANK YOU!!
     
  8. Dec 6, 2011 #7

    Mark44

    Staff: Mentor

    Yes, that's right.
     
  9. Dec 6, 2011 #8
    :) Thank you so much!
     
  10. Dec 6, 2011 #9
    I got confused again :(

    f(x) = 7cos2tsin6t

    I separated them like this: u= 7cos v =2t y =sin6t
    u'= -sin v'= 2 y'= cos6

    So far my derivative looks like this:

    f'(x) = -sin2tsin6t + 7cos2sin6t + 7cos2tcos6

    This is the farthest it can be simplified, right?
     
  11. Dec 6, 2011 #10

    eumyang

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    It probably would have been better if you started a new thread with the new problem. :wink:
    Is the function
    f(x) = 7 cos(2t) sin(6t)?
     
  12. Dec 6, 2011 #11
    That's probably true... Thank you for being my savior! :)

    It doesn't have parenthesis around them in my book, so I'm not sure at all... That's why I was so confused.
     
  13. Dec 6, 2011 #12
    Okay, I did it this way and it makes a lot more sense.

    f'(x)= -sin2tsin6t + 7cos2tcos6t

    I think that's all I can do with it, right?
     
  14. Dec 6, 2011 #13

    Dick

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    Simplification aside, that's not the derivative of 7*cos(2t)*sin(6t). Just like the derivative of cos(2t) is not -sin(2t). You have to use the chain rule.
     
  15. Dec 6, 2011 #14
    I see that now.

    Here goes, third times a charm (hopefully)...

    v' = -7sin2sin6t + 7cos2tcos6 + 0 * cos2tsin6t

    = -7sin2sin6t + 7cos2tcos6 (then you take out a 7, right?)

    v' = 7(-sin2sin6t)(cos2tcos6)

    I'm not sure if I'm still supposed to be adding them [(-sin2sin6t) + (cos2tcos6)] or not, but in the example earlier, getting rid of common things also got rid of the addition....

    Am I right now?

    (Thank you for helping :] )
     
  16. Dec 6, 2011 #15

    Dick

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    You are very welcome, but you are writing some serious nonsense here. Start with something simpler. What's the derivative of cos(2t)? Explain your answer. And use parentheses like eumyang suggested.
     
  17. Dec 6, 2011 #16
    I feel really ignorant now.

    derivative of cos(2t) = -2sin(2t)

    derivative of sin(6t) = 6cos(6t)

    derivative of 7 = 0
     
  18. Dec 6, 2011 #17

    Dick

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    That's great! Now use the product rule on 7 cos(2t) sin(6t). There won't be anything like sin(2) in there, will there?
     
  19. Dec 6, 2011 #18
    v' = -2sin(2t)sin(6t)*7 + cos(2t)*6cos(6t)*7 + 0

    v' = -14sin(2t)sin(6t) + 42cos(2t)cos(6t) [Took 14 out...]

    v' = 14(-sin(2t)sin(6t)) + 3(cos(2t)cos(6t)) [Do I continue to add them?]
     
  20. Dec 6, 2011 #19

    Dick

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    No. I think you should leave it as that. It's about as simple as it's going to get.
     
  21. Dec 6, 2011 #20
    Thank you.

    Would you mind if I ever messaged you with a question now and then?

    As you can probably tell, Calculus isn't exactly my strong suit.
     
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