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Chain Rule problem

  1. Sep 29, 2005 #1
    This is a problem that has stumped my entire class of Calc 1 students and two Calc 2 students.

    [tex] Find \frac {dy} {dx}[/tex]

    [tex]y = \frac {(2x+3)^3} {(4x^2-1)^8}[/tex]

    I know that the answer is (from the textbook, but I don't know how it got there)

    [tex] -\frac {2(2x+3)^2(52x^2+96x+3)} {(4x^2-1)^9}[/tex]

    I'll attempt to show my work in the next post.
  2. jcsd
  3. Sep 29, 2005 #2
    First I took the derivative using the chain rule of both the numerator and the denominator and applied them into the quotient rule.

    [tex]f' (2x+3)^3 = 6(2x+3)^2[/tex]
    [tex]g' (4x^2-1)^8 = 256x(4x-1)^7[/tex]

    Applying the quotient rule I get this

    [tex]\frac {6(4x^2-1)^8(2x+3)^2 - 256x(4x-1)^7(2x+3)^3} {[(4x^2-1)^8]^2}[/tex]

    I have no clue where they went from there...
  4. Sep 29, 2005 #3


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    g' is wrong, your coefficient in front of the seventh' power parenthesis should be 64, not 256.
  5. Sep 29, 2005 #4
    There should be a "squared" in the (4x - 1) bracket on the right of the top line of the quotient, then it just cancels.
  6. Sep 29, 2005 #5


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    As arildno said, your g' is wrong. Your denominator is [tex](4x^2-1)^{16}[/tex]
    You can divide all the terms by a [tex](4x^2-1)^7[/tex]. After that, it's just some factoring and multiplication in the numerator.
  7. Sep 29, 2005 #6
    I thought that because of the chain rule you have to take the derivative of the derivative of 4x^2 - 1 and multiply it by the others in sequence.

    God I hate the chain rule...

    I think I messed up g' more than I thought... You cannot touch the damn original function! :cry:
    Last edited: Sep 29, 2005
  8. Sep 29, 2005 #7


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    Did you get there now or still need help?
  9. Sep 29, 2005 #8
    Wait, I already did that, never mind...
  10. Sep 29, 2005 #9
    Ok am I on the right track here?

    [tex]\frac {6(4x^2-1)(2x+3)^2 - 64x(2x+3)^3} {(4x^2-1)^9}[/tex]
  11. Sep 29, 2005 #10
    You must take the derivative of 8x as per the chain rule and then multiply it into that term.

    Am I right?

    In that case I would have a 512 in front of that x, bleh
  12. Sep 29, 2005 #11


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    Yes, seems to be correct so far :smile:
  13. Sep 29, 2005 #12
    Was that to my last post? Because I somehow doubt it :(
  14. Sep 29, 2005 #13


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    It was in referrence to
    So you're doing ok so far.
  15. Sep 29, 2005 #14
    But that I don't get, as per the chain rule you must first get 8 from g', then 8x from what was in the brackets, then another 8 from 8x. Multiply them together and you get 512x, why is it still 64x?
  16. Sep 29, 2005 #15


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    You have a bit too many 8's then ...

    [tex]\left( {\frac{{\left( {2x + 3} \right)^3 }}
    {{\left( {4x^2 - 1} \right)^8 }}} \right)^\prime = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right)^8 - 64x\left( {4x^2 - 1} \right)^7 \left( {2x + 3} \right)^3 }}
    {{\left( {4x^2 - 1} \right)^{16} }} = \frac{{6\left( {2x + 3} \right)^2 \left( {4x^2 - 1} \right) - 64x\left( {2x + 3} \right)^3 }}
    {{\left( {4x^2 - 1} \right)^9 }}[/tex]

    One comes from the power (x^8 => 8x^7) and the other one from (4x²)' = 8x, which gives 8*8x = 64x.
  17. Sep 29, 2005 #16
    Yes, and then you must take the derivative of 8x and multiply that on because of the chain rule, right?

    I thought that as long as you can take a rational derivative with the chain rule, you should.
  18. Sep 29, 2005 #17
    [tex]8(4x^2-1)^7(8x)(8)[/tex] is how we do chain rule in class.

    Perhaps i'm horribly mistaken and i've forgotten part of the rules of the chain rule...
  19. Sep 29, 2005 #18


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    We're finished at that point, no more 8's to multiply...

    [tex]\left( {\left( {4x^2 - 1} \right)^8 } \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot \left( {4x^2 - 1} \right)^\prime = 8\left( {4x^2 - 1} \right)^7 \cdot 8x = 64x\left( {4x^2 - 1} \right)^7 [/tex]

    Ah no, you have an 8 extra. The derivative of 4x² is just 8x; not 8x*8 since (4x²)' = 4*(x²)' = 4*2x = 8x.
  20. Sep 29, 2005 #19
    Am I right so far in simplifying (using pascal's triangle and binomial theorem)

    [tex]\frac {6(4x^2-1)(2x+3)^2 - 64x(8x^3+36x^2+54x+27)} {4x^2-1)^9}[/tex]
  21. Sep 29, 2005 #20


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    Yes, that is correct. You also could've factored out [itex](2x+3)^2[/itex] first.
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