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Finding derivative of a square root - Quick question!

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Ok, working on a inverse function question, and I got stuck with something.

    Can someone explain the steps that makes this possible here. Something I am missing :(

    f(x) = √(x^3 + x^2 + x + 1)

    How is the inverse of the above function this...

    3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)


    2. Relevant equations



    3. The attempt at a solution

    I just looked at f(x) = √(x^3 + x^2 + x + 1) and went:

    1/2(3x^2 + 2x + 1)^-1/2

    Thats as far as I got...:(
     
  2. jcsd
  3. Nov 21, 2012 #2

    Dick

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    You aren't finding the inverse, you are taking the derivative. The part you are missing is the chain rule. (u^(1/2))'=(1/2)u^(-1/2)u'. Put u=x^3 + x^2 + x + 1. It's a special case of (f(g(x))'=f'(g(x))*g'(x).
     
  4. Nov 21, 2012 #3
    SORRY! Mistake... yes I am not finding the inverse. See below:

    f(x) = √(x^3 + x^2 + x + 1)

    How is the DERIVATIVE of the above function this...

    3x^2 + 2x + 1 / 2√(x^3 + x^2 + x + 1)
     
  5. Nov 21, 2012 #4

    Ray Vickson

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    What you wrote is not the inverse, but the derivative (but badly written). What you wrote reads as
    [tex] 3 x^2 + 2x + \frac{1}{2 \sqrt{x^3 + x^2 + x + 1}},[/tex]
    but perhaps you wanted to write
    [tex] \frac{3x^2 + 2x + 1}{2 \sqrt{x^3 + x^2 + x + 1}}.[/tex] In that case you should use parentheses, like this:
    (3x^2 + 2x + 1)/(2 √(x^3 + x^2 + x + 1)).

    The inverse of f(x) is much harder to get, as it would involve the solution of a 4th degree polynomial.

    RGV
     
  6. Nov 21, 2012 #5
    You are correct, sorry.
     
  7. Nov 21, 2012 #6

    LCKurtz

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    That isn't the inverse function. What it looks like is ##\frac{dy}{dx}## or ##f'(x)##.
    If you are calculating ##f'(x)## you aren't done. You need the chain rule (the derivative of the "inside"). Noting that you haven't actually stated the problem you are working on, let me guess that it has something to do with the derivative of the inverse function, not the inverse function itself. What is the relationship between the derivative of a function and the derivative of its inverse?

    [Edit] Wow! All those replies while I was typing. I must really be slow...
     
  8. Nov 21, 2012 #7

    Dick

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    I told you how. The numerator is u', the 2 in the denominator comes from (1/2) and √(x^3 + x^2 + x + 1) is u^(-1/2). Match them up with the parts of (u^(1/2))'=(1/2)u^(-1/2)u'. Chain rule.
     
  9. Nov 21, 2012 #8
    Thanks Dick, just got it!

    I got confused as to why the 3x^2 + 2x + 1 went on top.
     
  10. Nov 21, 2012 #9

    Mark44

    Staff: Mentor

    With over 600 posts here at PF you shouldn't be writing this:
    when you mean this:
    $$ \frac{3x^2 + 2x + 1}{2\sqrt{x^3 + x^2 + x + 1}}$$

    If using LaTeX is too much for you, at least use enough parentheses so that what you write can be clearly understood, like this:

    (3x^2 + 2x + 1) / (2√(x^3 + x^2 + x + 1))
     
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