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Math Help - find f'(3) and g'(4)

  1. #1
    Super Member bigwave's Avatar
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    find f'(3) and g'(4)

    1. If x[f(x)]^3+xf(x)=6 and f(3)=1, find f'(3)

    wasn't sure how to deal with this but thot if f(3)=1 then

    x[1]^3 + x[1]= 6
    x=3

    not sure how this helps nor how to get f'(X) to input f'(3)

    The answer is -\frac{1}{6}

    2. if [g(x)]^2 + 12x = x^2g(x) and g(4) = 12, find g'(4)

    there is no given answer to this one but still not sure how to deal with it
    Last edited by bigwave; March 12th 2010 at 03:44 PM. Reason: latex
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  2. #2
    Junior Member
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    Quote Originally Posted by bigwave View Post
    1. If x[f(x)]^3+xf(x)=6 and f(3)=1, find f'(3)

    wasn't sure how to deal with this but thot if f(3)=1 then

    x[1]^3 + x[1]= 6
    x=3

    not sure how this helps nor how to get f'(X) to input f'(3)

    The answer is -\frac{1}{6}

    2. if [g(x)]^2 + 12x = x^2g(x) and g(4) = 12, find g'(4)

    there is no given answer to this one but still not sure how to deal with it
    The first step to solving number one would be to implicitly derive the entire equation keeping in mind the product rule and the chain rule.

    After deriving you should have something like this:

    \left[ f\left( x \right) \right]^{3}\; +\; \left( 3\left[ f\left( x \right) \right]^{2}x \right)f'\left( x \right)\; +\; f\left( x \right)\; +\; x\left( f'\left( x \right) \right)\; =\; 0

    Then you collect terms and solve for f'(x):

    f'\left( x \right)\; =\; -\; \frac{f\left( x \right)\; +\; \left[ f\left( x \right) \right]^{3}}{3\left[ f\left( x \right) \right]^{2}x\; +\; x}

    Then you plug in the given values of x and f(x) to get

    f'\left( x \right)\; =\; -\; \frac{1}{6}

    then for number two you follow a similar method:
    first derive:

    2g\left( x \right)g'\left( x \right)\; +\; 12\; =\; 2xg\left( x \right)\; +\; x^{2}g'\left( x \right)

    solve for g'(x) then enter known values:

    g'\left( x \right)\; =\; \frac{2xg\left( x \right)\; -\; 12}{2g\left( x \right)\; -\; x^{2}}\; =\; \frac{21}{2}
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