Results 1 to 2 of 2

Math Help - Calculating second derivative

  1. #1
    Newbie
    Joined
    Oct 2007
    Posts
    11

    Calculating second derivative

    I have a function f(x) = sqrt(0.5*x^2 + 4) - 2

    I calculate the first derivative and get

    f'(x) = x / sqrt(0.5x^2 + 4)

    But I have a hard time trying to figure out the second derivative f''(x).

    How do I proceed?
    Last edited by Tikitac; October 13th 2007 at 09:13 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by Tikitac View Post
    I have a function f(x) = sqrt(0.5*x^2 + 4) - 2

    I calculate the first derivative and get

    f'(x) = 2x / sqrt(0.5x^2 + 4)

    But I have a hard time trying to figure out the second derivative f''(x).

    How do I proceed?
    Well, your first derivative isn't right anyway.

    Here's a helpful hint: Use fractions, it's a little clearer that way.
    f(x) = \sqrt{\frac{1}{2}x^2 + 4} - 2

    So
    f^{\prime}(x) = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{1}{2}x^2 + 4}} \cdot \left ( \frac{1}{2} \cdot 2x \right )

    Now to "pretty" this up a bit:
    f^{\prime}(x) = \frac{x}{2\sqrt{\frac{1}{2}x^2 + 4}}

    f^{\prime}(x) = \frac{x}{\sqrt{4 \cdot \frac{1}{2}x^2 + 4 \cdot 4}}

    f^{\prime}(x) = \frac{x}{\sqrt{2x^2 + 16}}

    For your second derivative, use the quotient rule:
    f^{\prime \prime}(x) = \frac{1 \cdot \sqrt{2x^2 + 16} - x \cdot \left ( \frac{1}{2} \cdot \frac{1}{\sqrt{2x^2 + 16}} \cdot (2 \cdot 2x) \right )}{2x^2 + 16}

    f^{\prime \prime}(x) = \frac{\sqrt{2x^2 + 16} - \frac{2x^2}{\sqrt{2x^2 + 16}}}{2x^2 + 16}

    f^{\prime \prime}(x) = \frac{\sqrt{2x^2 + 16} - \frac{2x^2}{\sqrt{2x^2 + 16}}}{2x^2 + 16} \cdot \frac{\sqrt{2x^2 + 16}}{\sqrt{2x^2 + 16}}

    f^{\prime \prime}(x) = \frac{2x^2 + 16 - 2x^2}{(2x^2 + 16)\sqrt{2x^2 + 16}}

    f^{\prime \prime}(x) = \frac{16}{\sqrt{(2x^2 + 16)^3}}

    The typical ways to write this would be
    f^{\prime \prime}(x) = \frac{16}{(2x^2 + 16)^{3/2}}
    or
    f^{\prime \prime}(x) = 16(2x^2 + 16)^{-3/2}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: November 25th 2011, 04:23 AM
  2. Calculating the second derivative of an integral
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: July 26th 2011, 01:46 PM
  3. Replies: 2
    Last Post: August 14th 2010, 06:45 AM
  4. Calculating Derivative in Several Variables
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: June 16th 2010, 01:54 PM
  5. Calculating the second derivative of a function
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 8th 2009, 07:42 AM

Search Tags


/mathhelpforum @mathhelpforum