Results 1 to 8 of 8

Math Help - 2nd Deriv to Pt Inflect (number 2)

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    19

    2nd Deriv to Pt Inflect (number 2)

    3x/(x+8).

    I need to use the 2nd derivative test to find the point(s) of inflection. I get confused about how to do derivatives of examples like this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by akuczma86 View Post
    3x/(x+8).

    I need to use the 2nd derivative test to find the point(s) of inflection. I get confused about how to do derivatives of examples like this.
    Quotient rule

    If f(x) = \frac{x}{y}

    then, f'(x) = \frac{x'y-y'x}{y^2}

    This should definitely be in your books/notes
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2010
    Posts
    19
    So the first derivative would be -3x/(x+8)^2?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    Quote Originally Posted by akuczma86 View Post
    So the first derivative would be -3x/(x+8)^2?
    NO.

    Its

    \frac {([\text{derivative of 3x}]\times (x+8)) - (3x \times [\text{derivative of} (x+8)])}{(x+8)^2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by akuczma86 View Post
    So the first derivative would be -3x/(x+8)^2? <<<<<< no
    f(x)=\frac{3x}{x+8}

    f'(x)=\frac{(x+8) \cdot 3 -3x \cdot 1}{(x+8)^2}

    Expand the bracket in the numerator, collect like terms.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2010
    Posts
    19
    f'(x)= 24/ (x^2 + 16x + 64

    f''(x)= (-24 -3x)/(x^4 + 2x^3 + 19x^2 + 48x + 36)

    Am I right here?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    It is OK for f'(x)

    BTW an easiest way in this case is f(x) = \frac{3x}{x+8} = 3 - \frac{24}{x+8}. The derivative of f is the same as the derivative of -\frac{24}{x+8} which is \frac{24}{(x+8)^2}

    The derivative of u^n(x) is n u'(x) u^{n-1}(x)

    Use this formula with u(x) = x+8 and n=-2 to get the derivative of \frac{1}{(x+8)^2} then multiply by 24
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by akuczma86 View Post
    f'(x)= 24/ (x^2 + 16x + 64

    f''(x)= (-24 -3x)/(x^4 + 2x^3 + 19x^2 + 48x + 36)

    Am I right here? <<<<<< I can't imagine how you got this result
    f'(x)= \frac{24}{x^2 + 16x + 64} then

    f''(x)= \frac{(x^2+16x+64) \cdot 0 - 24(2x+16)}{ (x^2 + 16x + 64)^2} = \boxed{\frac{ - 48x-384}{ x^4 + 32x^3 + 384x^2 + 2048x + 4096} }

    This result is correct but nearly useless. Compare with the method posted by running-gag:

    f'(x)= \frac{24}{x^2 + 16x + 64} = 24 (x+8)^{-2} then

    f'(x)= -48 (x+8)^{-3} \cdot 1 =\boxed{ -\frac{48}{(x+8)^3}}

    Both terms describe the same function!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Deriv
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 29th 2008, 01:16 AM
  2. deriv again
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 11th 2007, 04:11 PM
  3. log deriv.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 11th 2007, 08:40 AM
  4. deriv. of ln
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 8th 2007, 10:45 PM
  5. def. of deriv
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 7th 2007, 12:58 PM

Search Tags


/mathhelpforum @mathhelpforum