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The second derivative

  1. Nov 16, 2013 #1
    How do I find the second derivative of the function:

    f(x)= x^(2/3) (6-x)^(1/3)

    I have found the first derivative and checked my solution:

    ′()= 4− / ^(1/3) (6−)^(2/3)


    The final solution is supposed to be:

    ''()= -8 / ^(4/3) (6−)^(5/3)

    I know almost all the steps but I couldn't reach the final answer! Can you please help me?
    Thanks in advance!
     
  2. jcsd
  3. Nov 16, 2013 #2

    mfb

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    Staff: Mentor

    What did you get?
    Which differentiation rules do you know?
    It is impossible to see what you did wrong if you don't show your work.

    I think your f'(x) and f''(x) are missing brackets.
     
  4. Nov 16, 2013 #3

    lurflurf

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    let
    $$\mathrm{f}(x)=\mathrm{u}(x)\mathrm{v}(x) \\
    \text{with} \\
    \mathrm{u}(x)=x^{2/3} \\
    \mathrm{v}(x)=(6-x)^{1/3} $$
    By the product rule
    $$\mathrm{f}^{\prime \prime}(x)=\mathrm{u}^{\prime \prime}(x)\mathrm{v}(x)+2\mathrm{u}^{ \prime}(x)\mathrm{v}^{\prime }(x)+\mathrm{u}(x)\mathrm{v}^{\prime \prime}(x)$$
    what did you find for
    u'(x)
    u''(x)
    v'(x)
    v''(x)
    ?
     
  5. Nov 17, 2013 #4
    Where did you get this formula from? How did you wind up to this product rule?

    I got u'(x) = (2/3) x^(-1/3)
    u''(x) = (-2/9) x^(-4/3)

    v'(x) = -(1/3) < (6-x)^(-2/3) >
    v''(x) = -(2/9) < (6-x)^(-5/3) >


    using your formula it's difficult to simplify after this step

    < (-2/9) (x^(-4/3)) ((6-x)^(1/3)) > + < (-4/9) (x^(-1/3)) ((6-x)^(-2/3)) > + < (-2/9) (x^(2/3)) ((6-x)^(-5/3)) >
     
  6. Nov 17, 2013 #5

    mfb

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    Staff: Mentor

    lurfturf used the product rule both for f(x) and then for the f'(x) you get there.

    What are those < >?
    v'(x) has a wong prefactor.

    To simplify, you need exponentiation rules, especially a^(b+c)=...
     
  7. Nov 17, 2013 #6

    lurflurf

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    good, to simplify factor


    $$\mathrm{f}^{\prime \prime}(x)=\mathrm{u}^{\prime \prime}(x)\mathrm{v}(x)+2\mathrm{u}^{ \prime}(x)\mathrm{v}^{\prime }(x)+\mathrm{u}(x)\mathrm{v}^{\prime \prime}(x)
    \\
    =\left( x^{-2/3} \right)^{\prime \prime} (6-x)^{1/3}+2 \left( x^{-2/3}\right)^{\prime } \left( (6-x)^{-1/3}\right)^{\prime }+ x^{-2/3}\left( (6-x)^{-1/3}\right)^{\prime \prime}
    \\
    =\left( -\frac{2}{9} x^{-4/3}\right) \left( (6-x)^{1/3}\right)+2\left( \frac{2}{3} x^{-1/3} \right) \left( -\frac{1}{3}(6-x)^{-2/3}\right)+\left( x^{2/3}\right) \left( -\frac{2}{9}(6-x)^{-5/3} \right)
    \\
    -\frac{2}{9} x^{-4/3} (6-x)^{-5/3}((6-x)^2+2x(6-x)+x^2)
    \\
    =-\frac{2}{9}x^{-4/3}(6-x)^{-5/3}((6-x)+x)^2
    $$
    and so on
     
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