Results 1 to 2 of 2

Thread: derivative help #2

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    Santiago
    Posts
    517

    derivative help #2

    $\displaystyle
    \sqrt{sec\sqrt{x}}
    $

    $\displaystyle
    = (sec(x^\frac{1}{2}))^\frac{1}{2}
    $

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

    $\displaystyle
    = \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}
    $

    this the right path? thank u
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2007
    Posts
    108
    Quote Originally Posted by jvignacio View Post
    $\displaystyle
    \sqrt{sec\sqrt{x}}
    $

    $\displaystyle
    = (sec(x^\frac{1}{2}))^\frac{1}{2}
    $

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

    $\displaystyle
    = \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}
    $

    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.
    $\displaystyle f(x)=\sqrt{sec\sqrt{x}}=(sec(x^\frac{1}{2}))^\frac {1}{2}
    $

    Then $\displaystyle f'(x)=\frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \times (sec(x^\frac{1}{2}))'$
    $\displaystyle = \frac{1}{2}(sec(x^\frac{1}{2}))^\frac{-1}{2} \cdot sec(x)^\frac{1}{2}tan(x)^\frac{1}{2} \cdot \frac{1}{2}x^\frac{-1}{2}[$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Apr 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: Jan 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 23rd 2010, 06:33 AM
  4. Derivative Increasing ==> Derivative Continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Feb 23rd 2010, 10:58 AM
  5. Replies: 2
    Last Post: Nov 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum