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C3 Chain Rule Watch

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    Came across this question and my solution is different from the textbook answer.
    The curve C is defined by the equation:

    y = \dfrac{4}{x^2} + \dfrac{1}{2(1-x)^2}

    Show that there is only one stationary point and find its coordinates.

    I used chain rule then found dy/dx to be:

     (1-x)^{-3} - 8x^{-3}

    When I set this equal to 0 I proved there was one stationary point but my x-coordinate was 1/3 when the book had 2/3.

    Where have I went wrong?
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    (Original post by ozmo19)
    Came across this question and my solution is different from the textbook answer.
    The curve C is defined by the equation:

    y = \dfrac{4}{x^2} + \dfrac{1}{2(1-x)^2}

    Show that there is only one stationary point and find its coordinates.

    I used chain rule then found dy/dx to be:

     (1-x)^{-3} - 8x^{-3}

    When I set this equal to 0 I proved there was one stationary point but my x-coordinate was 1/3 when the book had 2/3.

    Where have I went wrong?
    your dy/dx is fine...

    1/( 1 - x ) 3 = 8/(x)3

    cube root both sides... 1/( 1 - x ) = 2/(x)

    *now rearrange & solve to get the book answer*
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    (Original post by the bear)
    your dy/dx is fine...

    1/( 1 - x ) 3 = 8/(x)3

    cube root both sides... 1/( 1 - x ) = 2/(x)

    *now rearrange & solve to get the book answer*
    I had done this but had done
    1/( 1 - x ) 3 = 1/8(x)3 by accident.

    Thank you!
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    (Original post by ozmo19)
    I had done this but had done
    1/( 1 - x ) 3 = 1/8(x)3 by accident.

    Thank you!
    we are here to help *
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    (Original post by the bear)
    we are here to help *
    Not sure how to go about this one:


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    A: it is a product rule with a little bit of chain too...

    u = e-x

    v = 4e2x - 2ex

    remember eax differentiates to aeax...

    B: alternatively you could multiply out the brackets then you would not need the product rule ...
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    (Original post by the bear)
    A: it is a product rule with a little bit of chain too...

    u = e-x

    v = 4e2x - 2ex

    remember eax differentiates to aeax...

    B: alternatively you could multiply out the brackets then you would not need the product rule ...
    Took du/dx as e^-x and not -e^-x - Thanks for the reminder
 
 
 
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