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Find dy of x/( sqrt (3x + 6) )

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    I am doing my review for an exam Friday and the review shows only this: "find dy"

    x/( sqrt(3x + 6) )

    The review gives me the answer which is:
    [(3x + 12)/2(3x + 6)^(3/2)]dx

    2. The attempt at a solution

    I started out by re-writing the sqrt as " (3x + 6)^1/2 " which I then proceeded to multiply the equation by the derivative of " (3x + 6) " and got:

    3x/(3x + 6)^(1/2).

    I'm stuck on how to proceed.
     
  2. jcsd
  3. Oct 23, 2012 #2

    Mentallic

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    Do you know the quotient rule for derivatives? Or equivalently, convert from

    [tex]\frac{x}{(3x+6)^{1/2}}[/tex]

    to

    [tex]x(3x+6)^{-1/2}[/tex]

    and use the product rule, setting [itex]u=x[/itex] and [itex]v=(3x+6)^{-1/2}[/itex]
     
  4. Oct 23, 2012 #3
    when you have functions like this, sometimes it's easier to write out your systems in general terms.

    rewrite your equation as f(x)/g(x), or f(x)g(x)^-1, if you prefer, where f(x)=x and g(x)=√(3x+6). so now find the derivative of this. do it piece by piece until you have your solution.
     
  5. Oct 23, 2012 #4

    SammyS

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    Hello HellRyu. Welcome to PF !

    The question should have read something like:
    Find dy, for [itex]\displaystyle y=\frac{x}{\sqrt{3x + 6}}\ .[/itex]​
    Of course, [itex]\displaystyle dy=\frac{dy}{dx} dx=y'\,dx[/itex]

    Basically, find y' and multiply by dx.

    Use the quotient rule with [itex]\displaystyle y=\frac{x}{(3x + 6)^{1/2}}\,,\ [/itex] or use the product rule with [itex]\displaystyle y=x(3x + 6)^{-1/2}\ .[/itex]
     
  6. Oct 24, 2012 #5
    Hi, and thank you! :)

    I started using the quotient rule which got me stuck at:

    [tex] (-3x^{2} + 3x + 6)/ ( (3x + 6)^{3/2} ) [/tex]
     
  7. Oct 24, 2012 #6

    SammyS

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    That's incorrect.

    Show some intermediate steps.
     
  8. Oct 24, 2012 #7
    O.K. I thought it was. I have:

    [tex] x/( (3x + 6)^{1/2} ) "f(x) / g(x)"[/tex]

    I then get [tex] "g(x)( f' (x) ) -( f(x)( g'(x) ) ) "[/tex]

    over

    [tex] "g(x)^{2} "[/tex]

    I plug my f(x)'s and g(x)'s with their derivatives

    [tex] (3x + 6)^{1/2} (1) - (x(3x) )[/tex]

    over

    [tex] (3x + 6)^{3/2} [/tex]
    Is this correct so far? Or did I make a mistake somewhere?
     
  9. Oct 24, 2012 #8

    MarneMath

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    No, you need to use the chain rule on the 2nd part.
     
  10. Oct 24, 2012 #9
    Ha! You guys are genius', thank you all so much! I figured it out.
     
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