1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivative of this function please?

  1. Mar 18, 2012 #1
    1. The problem statement, all variables and given/known data

    **NOTE**
    We are not allowed to use the quotient rule. :"(

    The function is

    y = ( (1 + square-root(x)) / (cube-root(x^2)) )^3


    3. The attempt at a solution

    I came up with a "simplified" expression.

    (1 + x^0.5)^3 * (x^-2)



    ***This is what I did, but it seems very wrong.***

    3(1 + x^.5)^2 * (0.5x^-0.5) + (1 + x^0.5)^3 * (-2x^-3)

    When trying to simply that I got into a mess. :(

    [3(1 + x^0.5)^2] / (2 * x^0.5) + [(1 + x^0.5)^3 * (-2x^-3)] / x^3

    I'm stuck now. :(

    Help please, thank you.
     
  2. jcsd
  3. Mar 18, 2012 #2

    Mark44

    Staff: Mentor

    You're missing a factor here.
    d/dx(f(x) * g(x)) = f'(x) * g(x) + g'(x) * f(x)
    In your work you used the chain rule correctly to get f'(x), but you omitted g(x). Your other term looks fine, since you have both g'(x) and f(x).
     
  4. Mar 18, 2012 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I guess this means$$
    \left ( \frac{1+\sqrt x}{x^{\frac 2 3}}\right)^3$$
    I added something you left out in red. You might not get a simple answer. You could maybe factor out some common factors.
     
  5. Mar 18, 2012 #4
    In the line where you apply the Product Rule, check your first term--there's a factor missing. Once that's fixed, you should be able to multiply one of the terms by something to get them both over the same denominator.

    You may also find it enlightening to try deriving the Quotient Rule in abstract form, that is:

    [tex]\frac{d}{dx}\frac{u}{v} = \frac{d}{dx}(uv^{-1}) = ...[/tex]

    That's a bit easier to work with, because there aren't as many exponents and stuff as your real problem. But if you're able to figure out how to do that in abstract form, you can follow the same procedure to manipulate your actual expression.
     
  6. Mar 18, 2012 #5
    Thanks guys, I'm working on it but I have a question.

    Is it possible for me to cancel out the cube-root and the cube and only apply the cube to the numerator of the fraction only and leave the bottom as x^2?
     
  7. Mar 18, 2012 #6
    Sure. That part of your simplification is fine, it's just the next step where things start going awry.
     
  8. Mar 18, 2012 #7
    >_<

    Lost again. :S

    Here's what I have so far.

    3(1 + x^.5)^2 * (0.5x^-0.5)x^-2 + (1 + x^0.5)^3 * (-2x^-3)

    I moved everything with a negative exponent down to simplify.

    3(1 + x^.5)^2 * (0.5x^-0.5)x^-2 + (1 + x^0.5)^3 * (-2x^-3)

    Ended up with.

    [3 * (1 + x^0.5)^2 * 0.5] / (x^0.5 * x^2) - (2 * (1 + x^0.5)^3) / x^3

    Do I have to find a common denominator?

    That seems scary. :(
     
  9. Mar 18, 2012 #8
    Ok. So far, we have:

    [tex]
    \frac{d}{dx}(\frac{1+\sqrt{x}}{\sqrt[3]{x^2}})^3\\
    = \frac{d}{dx}\frac{(1 + \sqrt{x})^3}{x^2}\\
    = \frac{d}{dx}(1+\sqrt{x})^3x^{-2}\\
    = 3(1+\sqrt{x})^2\frac{1}{2\sqrt{x}x^2} +(1+\sqrt{x})^3 \frac{-2}{x^3}
    [/tex]

    You're probably not going to be able to get it a whole lot simpler than that, but you should at least be able to factor out few common things, and then get the rest over a common denominator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivative of this function please?
  1. Functional Derivatives (Replies: 27)

Loading...