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Finding the derivative of a function with a radical

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    e1bbd57bb281ee733177266492e355.gif
    Find the differential
    2. Relevant equations
    Chain rule : dy/du=dy/du*du/dx
    Product rule: f(x)g'(x) + g(x)f'(x)
    3. The attempt at a solution
    I have tried to move the radical to the top of the equation by making it into an exponent (x^2+1)^-1/2. I then used the product rule and the chain ruleIand i got lost somewhere simplifying it beyond that.
     
  2. jcsd
  3. Feb 16, 2017 #2

    BvU

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    Hello Waffle, :welcome:

    PF culture pushes me to ask this :smile:

    Where were you when you got stuck ? In other words: show us how far you got and why you think you were lost at that point ...
     
  4. Feb 16, 2017 #3

    Ray Vickson

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    Show us what you got. Maybe your final,answer is correct, and maybe it isn't; we cannot tell if you will not show it to us.
     
  5. Feb 16, 2017 #4

    Mark44

    Staff: Mentor

    Please show us what you did. Your steps of using the product rule followed by the chain rule are the right approach.
    Also, do you need to find the differential or the derivative? They are related, but not the same.
     
  6. Feb 16, 2017 #5
    Sorry I meant to put derivative,
    This is where I got stuck at
    6x(x^2+1)^-1/2
    6x*d/dx(x^2+1)^-1/2+(x^2+1)^-1/2*6
    6x*1/2(x^2+1)^-3/2+(x^2+1)^-1/2*6
     
  7. Feb 16, 2017 #6
    Sorry I meant to put derivative,
    This is where I got stuck at
    6x(x^2+1)^-1/2
    6x*d/dx(x^2+1)^-1/2+(x^2+1)^-1/2*6
    6x*1/2(x^2+1)^-3/2+(x^2+1)^-1/2*6
     
  8. Feb 16, 2017 #7

    Ray Vickson

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    ##\frac{d}{dx} [(x^2+1)^{-1/2}] = (-1/2) (x^2+1)^{-3/2} \cdot \frac{d}{dx} (x^2+1) = -x(x^2+1)^{-3/2}##.
     
  9. Feb 16, 2017 #8
    Thank you
     
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