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Very simple question. Yes, I'm stupid.

  1. Nov 18, 2008 #1
    1. The problem statement, all variables and given/known data

    I am trying to solve this basic calculus problem, but I can't figure out the last step. It doesn't make sense to me. I know I'm just being an idiot and not thinking of it. This actually has little to do with calculus and more to do with basic algebra. Doh!


    2. Relevant equations

    The problems asks us to find the critical points of the function f(x) = x^3/5 * (4-x).

    3. The attempt at a solution

    I've gotten down to this point:
    f'(x) = (3(4-x) / 5x^2/5) - x^2/5.

    However, the solution to the problem goes on to this:
    f'(x) = (3(4-x) - 5x) / 5x^2/5

    I cannot for the life of me see how they reduced it down to that answer.

    Any help would be much appreciated!!!
     
  2. jcsd
  3. Nov 18, 2008 #2
    The highlighted 2 should be a 3. Then you can factor out the x2/5
     
  4. Nov 18, 2008 #3

    Mark44

    Staff: Mentor

    Your derivative is wrong.
    Given that [tex]f(x) = (4 - x) x^\frac{3}{5}[/tex]
    it's simpler to just multiply this out before taking the derivative.
    [tex]f(x) = 4x^\frac{3}{5} - x^\frac{8}{5}[/tex]
    [tex]f'(x) = \frac{12}{5}x^\frac{-2}{5} - \frac{8}{5}x^\frac{3}{5}[/tex]
    [tex]= x^\frac{-2}{5} (\frac{12}{5} - \frac{8}{5}x)[/tex]
    Now it's easy to find the critical point.
     
  5. Nov 18, 2008 #4
    I didn't think you could multiply it out?

    Don't you need to use the product rule to differentiate with respect to x?

    Also, there was as typo above. The x^2/5 ought to have been x^3/5.
     
  6. Nov 18, 2008 #5
    You can always simplify an expression before taking the derivative if this makes the calculation shorter... note that the derivative of x3/5 is 3/5 x-2/5
     
  7. Nov 18, 2008 #6
    I'm obviously rusty on my algebra here, but I greatly appreciate the help you guys are providing. All of it has really helped push me in the right direction. Also, I didn't know you could simplify before differentiating like that. Thank you for showing me that.

    I can see it if I simplify before hand; however, the professor is asking us to use the product rule. When I use the product rule, I can't see it, but I am getting closer.

    So here is where I'm at:

    I went from (3/5x^-2/5(4-x)) - x^3/5

    to 3/5x^-2/5((4-x) - 5/3x).

    This is close to the answer the book has of 3/5x^-2/5((4-x) - 5x).

    My question is, how did 5/3x turn into 5x?
     
  8. Nov 18, 2008 #7

    Mark44

    Staff: Mentor

    Please help us out here. If you're going to write expressions on a single line, use parentheses liberally. It's really hard to tell when you write -x^3/5 whether you are talking about [tex]\frac{-x^3}{5}[/tex] or [tex]-x^\frac{3}{5}[/tex], and that's a simple example.

    For the example I used, write it as -x^(3/5)
     
  9. Nov 18, 2008 #8
    I can see where the confusion is. My apologies.

    So here is where I'm at, making liberal use of parenthesis:

    I went from (((3/5)*(x^(-2/5)))*(4-x)) - (x^(3/5))

    to ((3/5)*(x^(-2/5)))*((4-x) - ((5/3)*x))


    This is close to the answer the book has of ((3/5)*(x^(-2/5)))*((4-x) - 5x).

    My question is, where did I go wrong by using ((5/3)*x), since the book has -5x? Or rather, how do I get to -5x from where I'm at?

    If there is any further confusion, please let me know, and I will try to clarify as best as possible.
     
  10. Nov 18, 2008 #9

    Mark44

    Staff: Mentor

    I am interpreting the answer you said in your previous post was the book's in this way:
    [tex]\frac{3}{5} x^ \frac{-2}{5} (4 - x - 5x)[/tex]
    This is not the right answer, and is in disagreement with what you said was the book's answer in your first post in this thread, namely
    f'(x) = (3(4-x) - 5x) / 5x^(2/5) (I added parentheses around the exponent.)

    This can be simplified to (12 - 3x - 5x)/(5x^(2/5)) = (12 - 8x)/(5x^(2/5))

    This is the same as (4/5)(3 - 2x)/(x^(2/5)), and is equal to what I got using two techniques.

    1. By multiplying first and then differentiating, I got
    [tex]f'(x) = \frac{4}{5} x^ \frac{-2}{5} (3 - 2x)[/tex]

    2. By using the product rule and then pulling out a common factor, I got
    [tex]f'(x) = \frac{3}{5} x^ \frac{-2}{5} (4 - x ) - x^\frac{3}{5}[/tex]
    [tex]= x^\frac{-2}{5} (\frac{3}{5} (4 - x ) - x)[/tex]
    [tex]= x^\frac{-2}{5} (\frac{12}{5} - \frac{3}{5}x - x)[/tex]
    [tex]= x^\frac{-2}{5} (\frac{12}{5} - \frac{8}{5}x)[/tex]
    [tex]= \frac{4}{5}x^\frac{-2}{5} (3 - 2x})[/tex]

    Either of my answers could also be written as
    [tex]f'(x) = \frac{8}{5}x^\frac{-2}{5} (1.5 - x})[/tex]

    The two expression I ended with in each case is preferred since you can do something with it, namely find where f'(x) = 0.

    Since I got the same answer using two different techniques, I'm pretty confident that I have the right answer. If your answer and/or the books answer can't be converted to look like mine, one or both of the answers you show are incorrect.
     
  11. Nov 18, 2008 #10
    Yes, I see what I am looking for. I was making a very small and simple mistake in the factoring, but it obviously had big consequences for my answer.

    I am not that particularly great at factoring, so this has really helped me.

    I am working to get better, and you guys have helped me progress. Thank you.
     
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