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Math Help - Question about a derivative problem

  1. #1
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    Question about a derivative problem

    The problem is simply

    f(x) = x^(3/2)


    using (f(x+h) - f(x)) / h and multiply with the conjugate I get

    (x+h)^3 - x^3
    /
    h(x+h)^(3/2) + z^(3/2)

    How do I get forward from this? I have been trying for hours now, but the algebra gets really ugly. Is there an easier way to reach the answer?

    Sorry I don't know how to write formulas in LaTex
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  2. #2
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    Re: Question about a derivative problem

    Consider \displaystyle \frac{ (x+h)^{\frac{3}{2}}- x^{\frac{3}{2} }}{h}= \frac{\sqrt{x+h}(x+h)- x\sqrt{x}}{h} = \frac{x\sqrt{x+h}+h\sqrt{x+h}- x\sqrt{x}}{h}
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  3. #3
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    Re: Question about a derivative problem

    Hello, leezgo!

    Differentiate: f(x) \,=\, x^{\frac{3}{2}}


    Using \frac{f(x+h) - f(x)}{h} and multiply with the conjugate,

    . . I get : . \frac{(x+h)^3 - x^3}{h\big[(x+h)^{\frac{3}{2}}+ x^{\frac{3}{2}}\big]} . Good!

    How do I get forward from this? . Keep going . . .

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

    . . . . . . . . . . . . . =\;\frac{h(3x^2 + 3xh + h^2)}{h\big[(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}\big]} \;=\;\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}}

    \lim_{h\to0}\frac{3x^2 + 3xh + h^2}{(x+h)^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\; \frac{3x^2}{x^{\frac{3}{2}} + x^{\frac{3}{2}}} \;=\;\frac{3x^2}{2x^{\frac{3}{2}}} \;=\; \frac{3x^{\frac{1}{2}}}{2}


    Therefore: . f'(x) \;=\;\tfrac{3}{2}x^{\frac{1}{2}}

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