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Homework Help: Finding the second derivative

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the exact value of f''(2) if f(x)=[tex]\sqrt{3x-4}[/tex]

    2. Relevant equations

    See above

    3. The attempt at a solution

    I've tried to use the product rule to differentiate.

    f= x(3x -4)[tex]^{\frac{1}{2}}[/tex]

    f'= (3x -4)[tex]^{\frac{1}{2}}[/tex] + [tex]\frac{3}{2}[/tex][tex]^{\frac{-1}{2}}[/tex]

    f''= [tex]\frac{3}{2}[/tex]x(3x -4)[tex]^{\frac{-1}{2}}[/tex] . [tex]\frac{3}{2}[/tex]x(3x -4)[tex]^{\frac{-1}{2}}[/tex] + (3x -4)[tex]^{\frac{1}{2}}[/tex] . [tex]\frac{-9}{4}[/tex]x(3x-4)[tex]^{\frac{-3}{2}}[/tex]

    = [tex]\sqrt{3x -4}[/tex] . [tex]\frac{-9x}{4\sqrt{(3x-4)^{3}}}[/tex]

    Now I am kind of stuck here. I've tried it a few times and I have no idea what I am doing wrong, some advice please?
     
  2. jcsd
  3. Feb 12, 2010 #2
    Use the chain rule instead
     
  4. Feb 12, 2010 #3
    Somehow I got the same answer :confused:

    f'(x) = [tex]\frac{3}{2}[/tex]x(3x -4)[tex]^{\frac{-1}{2}}[/tex]
    f''(x) = [tex]\frac{-9}{4}[/tex](3x -4)[tex]^{\frac{-3}{2}}[/tex]
     
  5. Feb 12, 2010 #4
    Your first derivative is incorrect. However, you have somehow ended up with the correct second derivative :S.

    f''(x) = [tex]\frac{-9}{4}[/tex](3x -4)[tex]^{\frac{-3}{2}}[/tex]


    [BTW: f'(x) = [tex]\frac{3}{2}[/tex](3x -4)[tex]^{\frac{-1}{2}}[/tex] ]
     
    Last edited: Feb 12, 2010
  6. Feb 12, 2010 #5

    Mark44

    Staff: Mentor

    Why are you writing your function like this? There is no factor of x outside the radical, so the product rule is not called for here. This is your function.
    [tex]f(x)=\sqrt{3x-4}[/tex]
     
  7. Feb 12, 2010 #6
    Sorry, I think the first derivative is this,

    f'(x) = [tex]\frac{1}{2}[/tex]x(3x -4)[tex]^{\frac{-1}{2}}[/tex]
     
  8. Feb 12, 2010 #7
    Please read your text book on chain rule. It is wrong to put an x in the front or a half for that matter.
     
  9. Feb 12, 2010 #8
    Ahh, whoops. The original equation to derive is actually [tex]f(x)=x\sqrt{3x-4}[/tex]!
     
  10. Feb 12, 2010 #9
    Ok, in that case you were right to use the product rule.

    f'(x) = [tex]\frac{3x}{2}(3x-4)^{-0.5}+(3x-4)^{0.5}[/tex]

    I leave the rest to you.
     
    Last edited: Feb 12, 2010
  11. Feb 12, 2010 #10
    Wait, how did you get that?
     
  12. Feb 12, 2010 #11
    Product rule.
    [tex]\frac{d[u(x)v(x)]}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}[/tex]
     
  13. Feb 12, 2010 #12
    Okay thank you, I've just realized that I've forgotten to multiply by the derivative of the brackets.
     
  14. Feb 12, 2010 #13
    Just in case you get stuck the answer is:

    [tex]3(3x-4)^{-0.5}-\frac{9}{4}x(3x-4)^{-1.5}[/tex]

    Though you should work this out by yourself. To verify.
     
  15. Feb 12, 2010 #14

    I got this, [tex]\frac{3}{2}(3x-4)^{-0.5}-\frac{9}{4}x(3x-4)^{-1.5}[/tex]

    But when finding the exact value of [tex]f''(2)[/tex] I had a bit of a problem as well.

    So far I have, [tex]f''(2)= \frac{-9}{4\sqrt{2}} + \frac{3}{2\sqrt{2}}[/tex].

    [tex]=\frac{-6\sqrt{2}}{4\sqrt{2}}[/tex]

    But the answer is [tex]\frac{3\sqrt{2}}{8}[/tex]
     
  16. Feb 13, 2010 #15
    That answer arises from my answer (i.e. the correct answer) go through your derivation again to check the mistakes.
     
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