Results 1 to 3 of 3

Math Help - differentiation again

  1. #1
    Member
    Joined
    Sep 2007
    Posts
    176

    differentiation again

    given x^{\frac 2 3} + y^{\frac 2 3} = a ^{\frac 2 3}
    find d^2 y / dx^2

    my working:
    dy/dx = -(\frac y x )^{1/3}
    d^2 y / dx^2 = {- \frac 1 3} (\frac y x)^ {- \frac 2 3}         \mbox{             }            {\frac{x \frac {dy}{dx} - y}{x^2}}
    = \frac {a^{\frac 2 3}}{3 x^{\frac 4 3} y ^{\frac 1 3}}
    <<<--------------- it's such a lengthy process! i cannot get the answer. is my method completely wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    It may be advantageous to resist algebraic manipulation before calculating the 2nd derivative. This allows the simpler product rule, rather than the cumbersome quotient rule.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,813
    Thanks
    703
    Hello, afeasfaerw23231233!

    Your work (so far) is correct . . .


    Given: . x^{\frac 2 3} + y^{\frac 2 3} \:= \:a ^{\frac 2 3} . . Find \frac{d^2y}{dx^2 }

    My working: . \frac{dy}{dx}\:=\:-\left(\frac{y}{x} \right)^{\frac{1}{3}}\;\;{\color{blue}\hdots \text{Yes!}}

    \frac{d^2y}{dx^2} \:=\: -\frac{1}{3}\left(\frac{y}{x}\right)^{-\frac{2}{3}}{\frac{x \frac{dy}{dx} - y}{x^2}}\;\;{\color{blue}\hdots\text{Right!}}

    Answer: . \frac {a^{\frac 2 3}}{3 x^{\frac 4 3} y ^{\frac 1 3}}
    Keep going . . .

    \text{We have: }\;\frac{d^2y}{dx^2} \;=\;-\frac{1}{3}\cdot\frac{x^{\frac{2}{3}}}{y^{\frac{2}  {3}}} \cdot\frac{x\left(-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}\right) - y}{x^2}

    . . . . . . . . . = \;-\frac{1}{3}\cdot\frac{x^{\frac{2}{3}}}{y^{\frac{2}  {3}}} \cdot\frac{-x^{\frac{2}{3}}y^{\frac{1}{3}} - y}{x^2} \;=\;-\frac{1}{3}\cdot\frac{x^{\frac{2}{3}}}{y^{\frac{2}  {3}}} \cdot\frac{-y^{\frac{1}{3}}\left(x^{\frac{2}{3}} + y^{\frac{2}{3}}\right)}{x^2}

    . . . . . . . . . = \;\frac{1}{3}\cdot\frac{\overbrace{x^{\frac{2}{3}} + y^{\frac{2}{3}}}^{\text{This is }a^{\frac{2}{3}}}}{x^{\frac{4}{3}}y^{\frac{1}{3}}}


    Therefore: . \frac{d^2y}{dx^2} \;=\;\frac{a^{\frac{2}{3}}}{3x^{\frac{4}{3}}y^{\fr  ac{1}{3}}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. Differentiation and partial differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 30th 2010, 10:16 PM
  3. Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 4th 2010, 10:45 AM
  4. Differentiation Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 25th 2010, 06:20 PM
  5. Differentiation and Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 6th 2009, 04:07 AM

Search Tags


/mathhelpforum @mathhelpforum