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Math Help - Piece Wise - Limits

  1. #1
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    Piece Wise - Limits

    Find the Derivative of
    Squareroot (x^3)

    By the limit defination of the derivative. Not the easy shortcut way. :/

    Can anyone point me in the right direction as to how I can start?

    I know the derivative is
    3/2 * x^1/2 (shortcut way) but can anyone show me the long way?
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  2. #2
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    \lim_{\Delta x\rightarrow 0}f(\frac{(\Delta x+x)-f(x)}{\Delta x})

    \frac{\sqrt{({\Delta x+x})^{3}}-\sqrt{{x}^{3}}}{\Delta x}

    Then keep solving
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  3. #3
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    Hello, EliteNewbz!

    Find the derivative by the limit definition of the derivative:

    . . . f(x) \:=\:\sqrt{x^3}
    We know the definition: . f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h}

    But I've never met any teacher who says this is a four-step procedure,
    . . and deliberately makes four steps. . (Am I the only one?)


    [1] Find f(x+h) . . . Replace x with x+h . . . and simplify.

    [2] Subtract f(x) . . . Subtract the original function . . . and simplify.

    [3] Divide by h . . . Factor and reduce.

    [4] Take the limit . . . Let h \to 0.



    We have: . f(x) \:=\:x^{\frac{3}{2}}


    [1] f(x+h) \:=\:(x+h)^{\frac{3}{2}}


    [2] f(x+h) - f(x) \;=\;(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}

    . . Multiply by \frac{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}

    . . f(x+h)-f(x)\;=\;\frac{(x+h)^{\frac{3}{2}} - x^{\frac{3}{2}}}{1}\cdot\frac{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} {(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{(x+h)^3 - x^3}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}

    . . f(x+h) - f(x) \;=\;\frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2h + 3xh^2 + h^3}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}


    [3] \frac{f(x+h) - f(x)}{h} \;=\;\frac{3x^2h + 3xh^2 + h^3}{h\left[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\right]} \;=\;\frac{h\left(3x^2 + 3xh + h^2\right)}{h\left[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\right]} . =\;\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}


    [4] \lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2 + 0 + 0}{x^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2}{2x^{\frac{3}{2}}} \;=\;\frac{3}{2}x^{\frac{1}{2}}


    Therefore: . \frac{d}{dx}\left(\sqrt{x^3}\right) \;=\;\frac{3}{2}x^{\frac{1}{2}}

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