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Find the derivative of the expression and simplify fully.

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data

    k(x) = x^2(2x^3+x)^1/3




    2. Relevant equations
    I used the product rule and the chain rule to get to the point that i did.



    3. The attempt at a solution

    dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)
    = (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2) I can't figure out how to solve further.
     
  2. jcsd
  3. Jul 15, 2011 #2
    A couple of questions:

    1) You state k(x) =... and then end up with dy/dx. Is it supposed to be y(x) at the beginning?

    2) Why are you doing this: "dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)"? There's no reason to add them together.

    Your solution is correct as far as I've been able to check out. This is how I would rewrite what you have there: dy/dx = (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2)

    You can also check your solution with wolframalpha.com.
     
  4. Jul 15, 2011 #3
    The solution is in my book, but i just don't understand how they got it.
     
  5. Jul 15, 2011 #4
    I don't know what you were adding at the beginning, but I agree with your answer.

    [itex]Dx[x^2(2x^3+x)^\frac{1}{3}][/itex]
    =

    [itex](x^2)Dx[(2x^2+x)^\frac{1}{3}] + (2x)(2x^3+x)^\frac{1}{3}[/itex]

    ..

    [itex]Dx[(2x^2+x)^\frac{1}{3}][/itex]

    let 1(x) = [itex]x^\frac{1}{3}[/itex]
    let 2(x) = u = (2x^3+x)

    [itex]Dx[mess] = 1'(u)2'(x)[/itex]
    =
    [itex]\frac{1}{3}(2x^3+x)^-{\frac{2}{3}}(6x^2+1)[/itex]

    ..
    [itex]Dx[x^2(2x^3+x)^\frac{1}{3}][/itex]
    =
    [itex](x^2)\frac{1}{3}(2x^3+x)^-\frac{2}{3}(6x^2+1) + (2x)(2x^3+x)^\frac{1}{3}[/itex]

    =

    [itex](\frac{x^2(6x+1)}{3(2x^3+x)^\frac{2}{3}})+(2x)(2x^3+x)^\frac{1}{3}[/itex]

    That's just how I'd write it, fewer terms and no negative exponents is my preference.
     
  6. Jul 15, 2011 #5

    in your last line, is that not supposed to be (6x^2 + 1)?
     
  7. Jul 15, 2011 #6
    Yes! Sorry.
     
  8. Jul 15, 2011 #7

    Mark44

    Staff: Mentor

    As already pointed out, since you are asked to differentiate k(x), your work should start with k'(x).
    Not mentioned is that you are using dy/dx improperly. dy/dx is the derivative of y with respect to x. To indicate that you are about to differentiate with respect to x, use the operator d/dx.

    Here is the corrected version of what you show above.

    k'(x) = d/dx[(x^2)(2x^3+x)^(1/3)] + [STRIKE]dy/dx(2x^3+x)^1/3(x^2)[/STRIKE]
    = (2x)(2x^3+x)^(1/3) + 1/3(2x^3+x)^(-2/3)(6x^2+1)(x^2)

    There is some additional work that can be done to put this in a more useful form, using factoring, but all the hard work (i.e., calculus) has been done.
     
  9. Jul 15, 2011 #8
    okay, i came to a fully simplified solution starting right after the above line, here it is:

    = 2x(2x^3+x)^1/3 + 1/3(1/(2x^3+x)^2/3(6x^2+1)(x^2)
    = 2x(2x^3+x)^1/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
    = 3(2x^3 + x)^2/3 (2x)(2x^3 + x)^1/3/3(2x^3 + x)^2/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
    = 6x(2x^3 + x) + 6x^4 + x^2/3(2x^3+ x)^2/3
    = 12x^4 + 6x^2 + 6x^4 + x^2/3(2x^3 + x)^2/3
    = 18x^4 + 7x^2/3(2x^3+x)^2/3
    = 1/3x^2(18x^2 + 7)(2x^3 + x)^-2/3
    It said to simplify fully so that is what i did.
     
  10. Jul 15, 2011 #9
    thank you.
     
  11. Jul 15, 2011 #10

    Mark44

    Staff: Mentor

    You're welcome!
     
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