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Math Help - Nasty derivative

  1. #1
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    Red face Nasty derivative

    Hello,
    I was wondering if someone could show me how to do this?
    Differentiate  a^{2} (a- a^{4} )^{3} Value of a=x^{-2}
    I can't figure it out
    Last edited by wolfhound; February 8th 2010 at 11:25 AM.
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    Hello,
    I was wondering if someone could show me how to do this?
    Differentiate  a^{2} (a- a^{4} )^{3} a=x^{-2}
    I can't figure it out
    Do you mean "differentiate both sides of this equation with respect to x"? If so, since the left side is just a constant with respect to x, and (x^n)'= nx^{n-1}, you will get 0= -2x^{-3}. That equation is never true however.

    If you mean "differentiate both sides of this equation with respect to a", then, since the right side is a constant with respect to a, and using the product rule and chain rule, (2a)(a- a^4)^3+ 3(a^2)(a- a^4)^2(1- 4a^3)= 0.

    If neither of those is what you mean, then I have no idea what you do mean.
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  3. #3
    Member mabruka's Avatar
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    The other posible thing going on is that we wants to differentiate both sides like when you do do normal substitution in integrals:



    u=x^2 \Longrightarrow  du=2xdx


    I am not sure either
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  4. #4
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    Quote Originally Posted by wolfhound View Post
    Hello,
    I was wondering if someone could show me how to do this?
    Differentiate  a^{2} (a- a^{4} )^{3} a=x^{-2}
    I can't figure it out
    a^{3} (a- a^{4} )^{3} = \frac{1}{x^2}  (lumped the a with a^2)

    Solve for \frac{da}{dx} via implicit differentiation ( a is clearly a function of x)

    \frac{d}{dx} \times (a^3 \times (a-a^4)^3)
    =
    3a^2(a-a^4)^3\frac{da}{dx} + 3a^3(a-a^4)^{2} (1-4a^3)\frac{da}{dx}
    =
    3a^2(a-a^4)^2(a-a^4+a(1-4a^3))\frac{da}{dx}
    =
    3a^2(a-a^4)^2(2a-5a^4)\frac{da}{dx}

    Also
    \frac{d}{dx}\frac{1}{x^2} = -2x^{-3}

    \Rightarrow

     \frac{da}{dx} = \frac{-2}{(3a^2(a-a^4)^2(2a-5a^4)x^3}
    Last edited by vince; February 8th 2010 at 11:35 AM.
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  5. #5
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    Smile

    Hello
    The value of a= x^-2
    yes I need to use the chain and product rule but all these exponents are mixing me up help,Please
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  6. #6
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    just great...now u tell us. oh well, my work above was consistent with the ambiguity of your intial post.
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  7. #7
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    Sorry I thought it made sense
    Please show me
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  8. #8
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    way too many little steps for typesetting through latex all the way through. at any rate, here is the setup:

    you want

    \frac{d}{dx}(\frac{1}{x^4}(\frac{1}{x^2}-\frac{1}{x^8})^3) =
    \frac{d}{dx}( \frac{1}{x^4}(\frac{x^6-1}{x^8})^3)
    =
    \frac{d}{dx}( \frac{(x^6-1)^3}{x^{28}}

    Differentiating that is easy but lengthy.
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  9. #9
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    Hi why raise x^8 to the power of 3,and not raise (x^6-1) to power of 3 aslo?
    is it easier to leave the top in brackets?
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  10. #10
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    Hi wolfhound,

    Vince has simply left the numerator "written" as a cube,
    since you can now apply the chain rule to it,
    or simply multiply out

    \left(x^6-1\right)\left(x^6-1\right)\left(x^6-1\right)

    Was it your idea to substitute a=x^{-2}
    in the beginning?

    Or was that requested by the question?

    You see, you can do things the short way using x,
    then manipulating the fraction to a simpler manageable one as Vince did.

    Or... use the messy substitution and get some good experience with the Chain Rule.

    Let me know the drive of the question,
    I can show you the way using the substitution if you want to play along with it.
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  11. #11
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    Smile

    Hello Archie
    It wasn't my Idea to use a=x^-2
    It was in the question,
    I cant seem to get this answer ( 28 -66x^6+48x^12-10x^18)/x^29

    Is it best to use the product+chain rule, or chain+quotient rule?
    Please help
    Thanks
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  12. #12
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    Hi wolfhound,

    Since you have a product after your substitution,
    you need to use a combination of product and chain rule.

    It's much faster to do it the way Vince showed.
    Nevertheless, here's what to do with the "a" substitution...

    \frac{d}{dx}\left(a^2\left(a-a^4\right)^3\right)=\frac{d}{dx}(uv)=v\frac{du}{dx  }+u\frac{dv}{dx}=v\frac{da}{dx}\frac{du}{da}+u\fra  c{da}{dx}\frac{dv}{da}

    u=a^2\ \Rightarrow\ \frac{du}{da}=2a

    v=\left(a-a^4\right)^3=w^3

    \frac{dv}{dw}=3w^2

    w=a-a^4\ \Rightarrow\ \frac{dw}{da}=1-4a^3

    \frac{dv}{da}=\frac{dv}{dw}\frac{dw}{da}=3w^2\left  (1-4a^3\right)=3\left(a-a^4\right)^2\left(1-4a^3\right)

    a=x^{-2}\ \Rightarrow\ \frac{da}{dx}=-2x^{-3}

    Putting all these together

    \frac{d}{dx}\left(a^2\left(a-a^4\right)^3\right)=v\frac{da}{dx}\frac{du}{da}+u\  frac{da}{dx}\frac{dv}{dw}\frac{dw}{da}

    We can write all of these terms using x to get the answer you are seeking.

    u=a^2=x^{-4}

    v=\left(a-a^4\right)^3=\left(x^{-2}-x^{-2(4)}\right)^3=\left(x^{-2}-x^{-8}\right)^3=\left(x^{-2}-x^{-8}\right) \left(x^{-2}-x^{-8}\right)\left(x^{-2}-x^{-8}\right)

    =\left(x^{-4}-2x^{-10}+x^{-16}\right)\left(x^{-2}-x^{-8}\right)=x^{-6}-3x^{-12}+3x^{-18}-x^{-24}

    \frac{da}{dx}=-2x^{-3}

    \frac{du}{da}=2x^{-2}

    \frac{dv}{dw}=3w^2=3\left(a-a^4\right)^2=3\left(a^2-2a^5+a^8\right)=3\left(x^{-4}-2x^{-10}+x^{-16}\right)

    \frac{dw}{da}=1-4a^3=1-4x^{-6}

    Therefore the derivative is

    \left(x^{-6}-3x^{-12}+3x^{-18}-x^{-24}\right)\left(-2x^{-3}\right)2x^{-2}+x^{-4}\left(-2x^{-3}\right) 3\left(x^{-4}-2x^{-10}+x^{-16}\right)\left(1-4x^{-6}\right)

    =-4x^{-11}+12x^{-17}-12x^{-23}+4x^{-29}-6x^{-7} \left(x^{-4}-4x^{-10}-2x^{-10}+8x^{-16}+x^{-16}-4x^{-22}\right)

    =4x^{-11}+12x^{-17}-12x^{-23}+4x^{-20}-6x^{-11}+36x^{-17}-54x^{-23}+24x^{-29}

    =-10x^{-11}+48x^{-17}-66x^{-23}+28x^{-29}

    =\frac{-10x^{18}+48x^{12}-66x^{6}+28}{x^{29}}
    Last edited by Archie Meade; February 10th 2010 at 03:18 AM.
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  13. #13
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    Thanks Archie,
    you really are my saviour !
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