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Homework Help: Which rule(s) is this using?

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Differentiate the following with respect to x;

    y = [itex]x^{2}[/itex][itex](x-1)^{1/2}[/itex]

    3. The attempt at a solution

    I have a solution to the problem that I will outline below, but my notes on this are confusing and i'm having trouble applying the method to another question. So if you can see the general rule that is being employed, it would really help me if you could point it out.

    Let u = [itex]x^{2}[/itex]

    Let v = [itex](x-1)^{1/2}[/itex]

    [itex]\frac{du}{dx}= 2x[/itex]

    [itex]\frac{dv}{dx}= \frac{1}{2}(x - 1)^{-1/2}[/itex]

    (that's all fine so far)

    [itex]\frac{dy}{dx}= \frac{x^{2}(x-1)^{-1/2}}{2} + 2x(x-1)^{1/2}[/itex]

    I have a simplified answer and I can see how to get there, but what rule does the above employ?

  2. jcsd
  3. Oct 28, 2013 #2


    User Avatar
    Homework Helper

    The solution is employing the product rule for differentiation where if you have y=uv


    dy/dx = v(du/dx_ + u(dv/dx)
  4. Oct 28, 2013 #3
    Thanks for the response,

    so am I correct in thinking that after splitting the function into u and v, the chain rule has been used to find dv/dx, the power rule can be used for du/dx and then from there it's plain sailing with the product rule?

    I definitely need more practice on breaking these problems down.
  5. Oct 28, 2013 #4


    Staff: Mentor

    Yes. In this case, the chain rule is very simple, since d/dx(x - 1) = 1.
  6. Oct 28, 2013 #5
    The problem I need to apply this method to is a little bit more complicated, but I can see how to do it.

  7. Oct 28, 2013 #6


    Staff: Mentor

    Here's another way to look at your problem.

    y = x2(x - 1)1/2
    dy/dx = x2 * d/dx[(x - 1)1/2] + d/dx(x2) * (x - 1)1/2
    = x2 * (1/2)(x - 1)-1/2 * d/dx(x - 1) + 2x * (x - 1)1/2
    = x2 * (1/2)(x - 1)-1/2 * 1 + 2x * (x - 1)1/2

    At each step along the way, I am postponing taking the derivative of something - this is signified by "d/dx( ... )", which means that I haven't actually taken the derivative of whatever is to its right.
  8. Oct 28, 2013 #7
    That's an interesting way of looking at it.

    I imagine it's much easier to look back through such a calculation and spot any potential error.
  9. Oct 28, 2013 #8


    Staff: Mentor

    Yes, because the work is shown inline rather than several lines up the page.
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