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Math Help - differentiation

  1. #1
    Senior Member furor celtica's Avatar
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    Thumbs down differentiation

    how do i get the gradient function of 1/((x-6)(x-2))? where does the power of -1 come in?
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  2. #2
    Senior Member Twig's Avatar
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    Hi!

    \frac{1}{(x-6)(x-2)}=\frac{1}{x^{2}-8x+12}

    \frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}

    The negative sign comes from  \frac{1}{x^{2}-8x+12} = (x^{2}-8x+12)^{-1}

    Then you multiply by the "inner function".
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  3. #3
    Senior Member furor celtica's Avatar
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    can you show all your work please? and how am i supposed to work from here to get the turning points? i sense a tedious fourth-degree equation...
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  4. #4
    Senior Member Twig's Avatar
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    Hi!

    -\frac{(2x-8)}{(x^{2}-8x+12)^{2}} is the derivative.

    Now factor out a 2.

    -2\cdot \frac{x-4}{x^{2}-8x+12} .

    Note that the function is not defined when  x = 6 \mbox{ or } x=2 (Why?)

    Now from the factored expression for the derivative we see that the derivative is zero when x=4 .

    To determine if it is a local maxima or local minima, you could use the second derivative test, or simply make a table.
    Note that if you plug in values slighly less than 4, you get a positive derivative, and values slight larger than 4 gives you a negative. So x=4 must be a local maxima.

    You also see that the derivative goes off to infinity as x approaches 2 and 6.
    Attached Thumbnails Attached Thumbnails differentiation-graph.jpg  
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  5. #5
    Senior Member furor celtica's Avatar
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    where did the power of two go in the beginning of this explanation?
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  6. #6
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    Hi furor celtica
    Quote Originally Posted by furor celtica View Post
    how do i get the gradient function of 1/((x-6)(x-2))? where does the power of -1 come in?
    As twig told you :
    \frac{d}{dx}\left(\frac{1}{x^{2}-8x+12}\right) = -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}

    To find the gradient, just substitute the desired value of x to \frac{dy}{dx}.

    Example : If you want to find the gradient at x= 1, subs. the value to \frac{dy}{dx}

    Quote Originally Posted by furor celtica View Post
    how am i supposed to work from here to get the turning points? i sense a tedious fourth-degree equation...
    where did the power of two go in the beginning of this explanation?
    To find turning points, set \frac{dy}{dx}=0. Then, you'll get :

    -\frac{(2x-8)}{(x^{2}-8x+12)^{2}}=0

    2x-8=0

    x=4
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