Results 1 to 9 of 9

Math Help - Finding the second derivative of a function

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    229

    Finding the second derivative of a function

    y = (x^3 - 2x)^4

    y' = 4(x^3 - 2x)^3 (3x^2 - 2)

    y'' = ?

    I'm having difficulty proceeding from there to get the second derivative. The answer is 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,786
    Thanks
    1570
    Quote Originally Posted by Archduke01 View Post
    y = (x^3 - 2x)^4

    y' = 4(x^3 - 2x)^3 (3x^2 - 2)

    y'' = ?

    I'm having difficulty proceeding from there to get the second derivative. The answer is 12(x^3 - 2x)^2 (11x^4 - 16x^2 + 4)
    Use the product rule.

    y'' = 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    229
    Quote Originally Posted by Prove It View Post
    Use the product rule.

    y'' = 4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}
    Hi again, using the product rule I get;

    y'' = 4[3(x^3 - 2x)^2 (3x^2 - 2) (3x^2 - 2) + (x^3 - 2x)^3 (6x)

    I'm not sure if it's correct or not but regardless I don't know how to simplify the above expression.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,786
    Thanks
    1570
    Quote Originally Posted by Archduke01 View Post
    Hi again, using the product rule I get;

    y'' = 4[3(x^3 - 2x)^2 (3x^2 - 2) (3x^2 - 2) + (x^3 - 2x)^3 (6x)

    I'm not sure if it's correct or not but regardless I don't know how to simplify the above expression.
    Yes, now you can take out common factors

    4\left\{(x^3 - 2x)^3 \frac{d}{dx}(3x^2 - 2) + (3x^2 - 2)\frac{d}{dx}[(x^3 - 2x)^3]\right\}

     = 4[6x(x^3 - 2x)^3 + 3(3x^2 - 2)(3x^2 - 2)(x^3 - 2x)^2]

    Now take out common factors.

     = 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]

     = 12(x^3 - 2x)^2[2x^4 - 4x^2 + 9x^4 - 12x^2 + 4]

     = 12(x^3 - 2x)^2(11x^4 - 16x^2 + 4)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    229
    Quote Originally Posted by Prove It View Post

     = 12(x^3 - 2x)^2[2x(x^3 - 2x) + (3x^2 - 2)^2]
    What happened to the 6x?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,786
    Thanks
    1570
    Quote Originally Posted by Archduke01 View Post
    What happened to the 6x?
    I took out a common factor of 3, which leaves the 2x.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2009
    Posts
    229
    Quote Originally Posted by Prove It View Post
    I took out a common factor of 3, which leaves the 2x.
    The common factor of 3 was only common with 6x? Why did you take out a factor 3 specifically?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,786
    Thanks
    1570
    Quote Originally Posted by Archduke01 View Post
    The common factor of 3 was only common with 6x? Why did you take out a factor 3 specifically?
    Read it carefully.

    The first term is 6x(x^3 - 2x)^3 has a factor of 3.

    You will see that the second term is 3(3x - 2)^2(x^3 - 2x)^2, which also has a factor of 3.

    Therefore I can take out 3 as a common factor.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Oct 2009
    Posts
    229
    Quote Originally Posted by Prove It View Post
    Read it carefully.

    The first term is 6x(x^3 - 2x)^3 has a factor of 3.

    You will see that the second term is 3(3x - 2)^2(x^3 - 2x)^2, which also has a factor of 3.

    Therefore I can take out 3 as a common factor.
    Oh right, I see it. I missed it originally because I had already taken out the 3 in the second term by accident.

    Thanks for your help; I understand it now. And sorry for taking up your time with questions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the derivative of the function
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 24th 2011, 07:02 PM
  2. Replies: 4
    Last Post: December 13th 2010, 01:17 PM
  3. Finding the derivative of this function:
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 06:57 PM
  4. Finding derivative of function
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 18th 2009, 02:53 PM
  5. Finding a Derivative of a Function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 28th 2009, 10:24 PM

Search Tags


/mathhelpforum @mathhelpforum