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How to solve the equation with square roots

  1. May 2, 2009 #1
    1. The problem statement, all variables and given/known data
    I am having trouble to find out what rule to use when solving this equation:
    y=5x^2 square root of (2x^3)/15x^7 square root x

    How do you right square roots on the keyboard. As you can see I am quit new to this forum


    2. Relevant equations
    I have tried to use the chain rule and the quotiant rule. But when I use a different rule I get a different answer.



    3. The attempt at a solution
     
  2. jcsd
  3. May 3, 2009 #2
    Re: Differentiation

    Is this the correct equation? (you can see the markup by quoting me)

    [tex]y=5x^2 \sqrt{\frac{2x^3}{15x^7}\sqrt{x}}[/tex]
     
  4. May 3, 2009 #3
    Re: Differentiation

    No
    I think it is [tex]y=5x^2 {sqrt{2x^3}}\15x^7{sqrt{x}}[/tex]
     
  5. May 3, 2009 #4
    Re: Differentiation

    This is what I mean sorry for that

    [tex]y=5x^2\sqrt{2x^3}{frac{15x^7}\sqrt{x}} tex][/QUOTE]
     
  6. May 3, 2009 #5
    Re: Differentiation

    You can differentiate x to a fractional powers the same way you can differentiate x to an integral power. Try simplifying it first.
     
  7. May 3, 2009 #6
    Re: Differentiation

    Thanks anyway I cann't get the question right. The 5x^2 and sqrt(2x^3) are on the top
    and 15x^7 and sqrt(x) are on the bottom. lets try that.
     
  8. May 3, 2009 #7

    HallsofIvy

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    Re: Differentiation

    So
    [tex]\frac{5x^2\sqrt{2x^3}}{15x^7\sqrt{x}}[/tex]?
    As qntty suggested, you can write the square root as a fractional power: [itex]\sqrt{x}= x^{1/2}[/itex].

    Your expression is
    [tex]\frac{5\sqrt{2}x^2x^{3/2}}{15x^7x^{1/2}}= \frac{5\sqrt{2}x^{2+3/2}}{15x^{7+ 1/2}}= \frac{5\sqrt{2}x^{7/2}}{15x^{15/2}}[/tex]
    [tex]= \frac{\sqrt{2}}{3}x^{7/2}x^{-15/2}= \frac{\sqrt{2}}{3}x^{-4}[/tex]
    That should be easy to differentiate.
     
    Last edited: May 4, 2009
  9. May 3, 2009 #8
    Re: Differentiation

    Thank you very much. Can you just tell me what rule you used, so I can apply it to other equations.
     
  10. May 4, 2009 #9

    HallsofIvy

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    Re: Differentiation

    I used a number of "rules". I used the fact that [itex]\sqrt{x}= x^{1/2}[/itex], I used the laws of exponents to reduce the problem to a single power of x, and I used the power rule to differentiate.
     
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