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Math Help - derivative help #2

  1. #1
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    derivative help #2

     <br />
\sqrt{sec\sqrt{x}}<br />

     <br />
= (sec(x^\frac{1}{2}))^\frac{1}{2}<br />

     <br />
= \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2}<br />

     <br />
= \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}<br />

    this the right path? thank u
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  2. #2
    Member
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    Quote Originally Posted by jvignacio View Post
     <br />
\sqrt{sec\sqrt{x}}<br />

     <br />
= (sec(x^\frac{1}{2}))^\frac{1}{2}<br />

     <br />
= \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2}<br />

     <br />
= \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}<br />

    this the right path? thank u
    I think your answer is correct. It's just the way you present it. It doesn't make sense to put all those equal signs because they are clearly are not all equal.
    f(x)=\sqrt{sec\sqrt{x}}=(sec(x^\frac{1}{2}))^\frac  {1}{2}<br />
    Then f'(x)=\frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \times (sec(x^\frac{1}{2}))'
    = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}[
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