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Math Help - Derivative/chain rule

  1. #1
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    Derivative/chain rule

    f(x) = \frac{1}{x\sqrt{5-2x}}
    <br />
f'(x) = (x\sqrt{5-2x})^{-1} \\<br />
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\<br />
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\<br />
= -(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)') \\<br />
= -(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))) \\<br />
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\<br />
= -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)}<br />

    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:

    = \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\<br />
\\<br />
\\<br />
= \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; July 13th 2012 at 07:06 PM.
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  3. #3
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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
    = -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)}

    After I simplify, I get this:

    = \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

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

    =\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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