Results 1 to 4 of 4

Thread: Finding the derivative using the definition (have answer)

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    92

    Finding the derivative using the definition (have answer)

    Find the derivative using the definition (NOT shortcuts, we must use definition).

    y=1/sqrt (x)

    My friend and I got it down to x^(-1) - x^(-3/2)
    but the answer is: -1/2x^(3/2)

    How did they get that? (not using shortcuts)
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,800
    Thanks
    2829
    Awards
    1
    $\displaystyle \frac{{\frac{1}
    {{\sqrt {x + h} }} - \frac{1}
    {{\sqrt x }}}}
    {h} = \frac{{\sqrt x - \sqrt {x + h} }}
    {{h\sqrt x \left( {\sqrt {x + h} } \right)}} = \frac{{ - h}}
    {{h\sqrt x \left( {\sqrt {x + h} } \right)\left( {\sqrt x + \sqrt {x + h} } \right)}}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    $\displaystyle \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} $

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}$

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

    $\displaystyle = \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Dec 2008
    Posts
    92
    Quote Originally Posted by Random Variable View Post
    $\displaystyle \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} $

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}$

    $\displaystyle = \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}$

    $\displaystyle = \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}$

    THANK YOU!! You're a genius lol. I'm so happy I know how to do this now. :]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 23rd 2011, 09:15 AM
  2. Replies: 2
    Last Post: Jan 13th 2010, 08:26 AM
  3. Finding a derivative by definition
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 10th 2009, 08:09 AM
  4. Replies: 2
    Last Post: Nov 6th 2009, 02:51 PM
  5. Finding derivative from limit definition
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 29th 2009, 08:40 PM

Search Tags


/mathhelpforum @mathhelpforum