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Math Help - derivative of 1/[sqrt[x]: proving with derivative definition

  1. #1
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    derivative of 1/[sqrt[x]: proving with derivative definition

    i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

    placed the function in the definition

    1/sqrt[(x+h)] - 1/sqrt[x]
    ----------------------------
    h

    [(x+h)^-.5] - [x^-.5]
    ------------------------
    h

    apply conjugate:
    [(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
    ------------------------
    h * [(x+h)^-.5] + [x^-.5]


    im lost here on out.
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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post
    i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

    placed the function in the definition

    1/sqrt[(x+h)] - 1/sqrt[x]
    ----------------------------
    h

    [(x+h)^-.5] - [x^-.5]
    ------------------------
    h

    apply conjugate:
    [(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
    ------------------------
    h * [(x+h)^-.5] + [x^-.5]


    im lost here on out.
    \frac{\frac{1}{\sqrt{x+h}} - \frac{1}{\sqrt{x}}}{h} = \frac{\sqrt{x} - \sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} = \frac{x-x-h}{h\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})}   = \frac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x} + \sqrt{x+h})} \to \frac{-1}{\sqrt{x}\sqrt{x}(\sqrt{x} + \sqrt{x})} = \frac{-1}{2x^{\frac{3}{2}}}
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  3. #3
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    Quote Originally Posted by Evan.Kimia View Post
    i need to solve 1/sqrt[x] using the limit method. i checked it first using the various rules and got 1/2x^3/2. heres what i did:

    placed the function in the definition

    1/sqrt[(x+h)] - 1/sqrt[x]
    ----------------------------
    h

    [(x+h)^-.5] - [x^-.5]
    ------------------------
    h

    apply conjugate:
    [(x+h)^-.5] - [x^-.5] * [(x+h)^-.5] + [x^-.5]
    ------------------------
    h * [(x+h)^-.5] + [x^-.5]


    im lost here on out.

    Good idea to use the conjugate but you have to push forward a little more:

    \lim_{h\to 0}\,\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{h\to 0}\,\frac{\frac{1}{x+h}-\frac{1}{x}}{h\left(\frac{1}{\sqrt{x+h}}+\frac{1}{  \sqrt{x}}\right)} =\lim_{h\to 0}\,\frac{-h}{h\left(\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}\  right)x(x+h)}=\lim_{h\to 0}\,\frac{-1}{\left(\frac{1}{\sqrt{x+h}}+\frac{1}{\sqrt{x}}\r  ight)x(x+h)} =-\frac{1}{\frac{2}{\sqrt{x}}x^2}=-\frac{1}{2x^{3/2}}=-\frac{1}{2x\sqrt{x}}

    Tonio
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