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Math Help - I really need help on derivatives

  1. #1
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    Unhappy I really need help on derivatives

    Part I
    a.) find the slope of the tangent lineto the graph of the function f at point (x_1,f(x_1)). b.) Find the points on the graph where the tangent line is horizontal and use these points to sketch the graph.
    f(x) = 2x^3 - 3x^2

    Part II
    Find the indicated derivative:
    D_x(\frac{1}{x^2} - x)

    Hey can you use the
    \lim_{h\to_ 0} \frac{f(x+h) - f(x)}{h} to solve the formula? coz i can really get it there

    Part III
    Find \frac{dy}{dx}:

    y= \frac{4}{2x-5}
    Last edited by ^_^Engineer_Adam^_^; December 1st 2006 at 03:09 AM.
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  2. #2
    TD!
    TD! is offline
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    Unless you are asked to use the limit definition of the derivative, why not just use the rules/properties?
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Part I
    a.) find the slope of the tangent lineto the graph of the function f at point (x_1,f(x_1)). b.) Find the points on the graph where the tangent line is horizontal and use these points to sketch the graph.
    f(x) = 2x^3 - 3x^2
    Is this correct?
    a.) f(x_1) = 6x_1^2 - 6x_1

    Is my graph correct?
    (1,-1) & (0,0) are the horizontal tangent lines
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Is this correct?
    a.) f(x_1) = 6x_1^2 - 6x_1

    Is my graph correct?
    (1,-1) & (0,0) are the horizontal tangent lines
    Looks good to me.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Part II
    Find the indicated derivative:
    D_x(\frac{1}{x^2} - x)
    \frac{d}{dx} \left ( \frac{1}{x^2} - x \right ) = -\frac{2}{x^3} - 1

    -Dan
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Part III
    Find \frac{dy}{dx}:

    y= \frac{4}{2x-5}
    There are two ways to look at this:
    1) The quotient rule:
    \frac{dy}{dx} = \frac{0 \cdot (2x-5) - 4(2)}{(2x-5)^2} = -\frac{8}{(2x-5)^2}

    2) y = 4(2x-5)^{-1} = 4(-1)(2x-5)^{-2} \cdot 2 = -8(2x-5)^{-2} = -\frac{8}{(2x-5)^2}

    -Dan
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