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Thread: Derivative/chain rule

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    Derivative/chain rule

    $\displaystyle f(x) = \frac{1}{x\sqrt{5-2x}}$
    $\displaystyle
    f'(x) = (x\sqrt{5-2x})^{-1} \\
    = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\
    = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\
    = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\
    = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\
    = -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\
    = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)}
    $

    Have I done anything wrong so far?
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  2. #2
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    Re: Derivative/chain rule

    After I simplify, I get this:

    $\displaystyle = \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\
    \\
    \\
    = \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}$

    But the solution says I'm wrong...what have I done?
    Last edited by daigo; Jul 13th 2012 at 06:06 PM.
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    Re: Derivative/chain rule

    Oh wait, I see it now...the very first negative sign at the very beginning...I completely missed that this entire time and I couldn't see it...wow
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    Re: Derivative/chain rule

    Quote Originally Posted by daigo View Post
    $\displaystyle = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)}$

    After I simplify, I get this:

    $\displaystyle = \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)}$
    You forgot the negative sign that was in front. That should be

    $\displaystyle f'(x) = -\frac{(5-2x)^{1/2}}{x^2(5-2x)} + \frac{x(5-2x)^{-1/2}}{x^2(5-2x)}$

    $\displaystyle =\frac1{x(5-2x)^{3/2}} - \frac1{x^2(5-2x)^{1/2}}$

    This is the correct derivative. Naturally, your textbook might present it in a different form.
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