Results 1 to 6 of 6

Math Help - 2nd Derivative

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    21

    2nd Derivative

    How do you solve the 2nd derivative of this problem?
    [LaTeX ERROR: Convert failed]
    I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by sgonzalez90 View Post
    How do you solve the 2nd derivative of this problem?
    [LaTeX ERROR: Convert failed]
    I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.
    Let g(x) = f'(x)

    Let u = -48x and so u' = -48

    Let v = (x^2+12)^2 and using the chain rule we get v' = 2(x^2+12) \times 2x = 4x(x^2+12) = 4x^3+48x

    The quotient rule says: g'(x) = f''(x) = \frac{u'v - uv'}{v^2}

    Can you take it from there?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by sgonzalez90 View Post
    How do you solve the 2nd derivative of this problem?
    f'(x)=-48x/(x^2 +12)^2
    I know you have to use the quotient rule and chain rule. I just don't know how to get to the answer correctly. I keep screwing up, so if you could just walk me through this it would be great.

    i prefer to write it like this

    f'(x)=-48x(x^2+12)^{-2}

    f''(x)=-48(x^2+12)^{-2}+(-2)(2x)(x^2+12)^{-3}(-48x)

    or use

    f'(x)=\frac{-48x}{(x^2+12)^2}

    f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}

    you simplify
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2009
    Posts
    21
    Quote Originally Posted by e^(i*pi) View Post
    Let g(x) = f'(x)

    Let u = -48x and so u' = -48

    Let v = (x^2+12)^2 and using the chain rule we get v' = 2(x^2+12) \times 2x = 4x(x^2+12) = 4x^3+48x

    The quotient rule says: g'(x) = f''(x) = \frac{u'v - uv'}{v^2}

    Can you take it from there?
    That answer is incorrect

    The answer is: [LaTeX ERROR: Convert failed]
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    21
    Quote Originally Posted by Amer View Post
    i prefer to write it like this

    f'(x)=-48x(x^2+12)^{-2}

    f''(x)=-48(x^2+12)^{-2}+(-2)(2x)(x^2+12)^{-3}(-48x)

    or use

    f'(x)=\frac{-48x}{(x^2+12)^2}

    f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}
    Okay, like i said. I already know that this is what I'm supposed to get. But How do i simplify this to get the answer I just posted?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2009
    Posts
    53
    <br />
f''(x)=\frac{-48(x^2+12)^{2}-(-48x)(2)(2x)(x^2+12)}{(x^2+12)^{4}}<br />

    Factor out (x^2 + 12):

    <br />
f''(x)=\frac{-48(x^2+12)-(-48x)(2)(2x)}{(x^2+12)^{3}}<br />

    Expand:

    <br />
f''(x)=\frac{-48x^2 - 576 + 192x^2}{(x^2+12)^{3}}<br />

    You get:

    <br />
f''(x)=\frac{144x^2 - 576}{(x^2+12)^{3}}<br />
    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: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 06:33 AM
  4. Derivative Increasing ==> Derivative Continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 23rd 2010, 10:58 AM
  5. Replies: 2
    Last Post: November 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum