Results 1 to 4 of 4

Math Help - Trouble Finding Derivative

  1. #1
    Junior Member
    Joined
    Sep 2011
    Posts
    27

    Trouble Finding Derivative

    I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

     \frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]

    Also, in case it matters, d=24 and w=6, but in the example, the derivative is found with respect to the variables rather than their respective values.

    My attempt:

     [(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]

     [(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

     [(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

    From this point, I could do stuff like factor out a 2(x+d), but it looks like mine won't be very easy to solve for zero.

    This is what the derivative is supposed to be, which is easier to solve for zero:

     (x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)

    What am I doing incorrectly?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Trouble Finding Derivative

    Quote Originally Posted by Algebrah View Post
    I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

     \frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]

    Also, in case it matters, d=24 and w=6, but in the example, the derivative is found with respect to the variables rather than their respective values.

    My attempt:

     [(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]

     [(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

     [(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

    From this point, I could do stuff like factor out a 2(x+d), but it looks like mine won't be very easy to solve for zero.

    This is what the derivative is supposed to be, which is easier to solve for zero:

     (x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)

    What am I doing incorrectly?

    Thanks!
    1. You have introduced and extra factor of two in the first term erroneously.

    2. The second term should be:

     (x+d)^2 \frac{d}{dx}\left( \frac{w^2}{x^2}+1\right) =(x+d)^2 \frac{(-2)w^2}{x^3}

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470

    Re: Trouble Finding Derivative

    Quote Originally Posted by Algebrah View Post
    I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

     \frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]

    Also, in case it matters, d=24 and w=6, but in the example, the derivative is found with respect to the variables rather than their respective values.

    My attempt:

     [(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]
    Where did that "(1+1)"? Surely you didn't differentiate "(x+ d)" and get "(1+ 1)"? "d" is a constant. The derivative of x+ d with respect to x is 1. Also you appear to have tried to write \frac{w^2}{x^2}+ 1= \frac{w^2+ x^2}{x^2} and use the quotient rule- but you have done that incorrectly. Better is to write \frac{w^2}{x^2}+ 1= w^2x^{-2}+ 1 so the derivative is -2w^2x^{-3}= \frac{-2w^2}{x^3}.


     [(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

     [(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]

    From this point, I could do stuff like factor out a 2(x+d), but it looks like mine won't be very easy to solve for zero.

    This is what the derivative is supposed to be, which is easier to solve for zero:

     (x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)

    What am I doing incorrectly?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2011
    Posts
    27

    Re: Trouble Finding Derivative

    Perhaps if I hadn't neglected the \frac{d}{dx} in front, things would've come together a lot quicker! Lesson learned.

    Thanks a lot for the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative trouble # 2
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 25th 2011, 05:18 PM
  2. Derivative trouble # 1
    Posted in the Calculus Forum
    Replies: 12
    Last Post: May 25th 2011, 08:45 AM
  3. Replies: 3
    Last Post: April 25th 2010, 01:34 AM
  4. Having trouble with derivative..please help.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 16th 2010, 01:31 PM
  5. derivative trouble
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 15th 2007, 06:12 PM

/mathhelpforum @mathhelpforum