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Math Help - Answer check please

  1. #1
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    Answer check please

    Hi would somebody be able to check my answers to some questions I have, i'll just post one at a time as they take up a bit of room:

    Q: Obtain f''(0.3) for each of the following functions.

    i) f(x) = \arccos x

    A:

    f(x) = \arccos x
    f'(x) = -(1-x^2)^ \frac{-1}{2}

    Sub z = 1-x^2
    y = -z^\frac{-1}{2}

    \frac{dy}{dz} = \frac{1}{2}(1-x^2)^\frac{-3}{2}

    f''(x) = \frac{-1}{\sqrt{(1-x^2)^3}}

    f''(0.3) = 1.152
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  2. #2
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    I've done the first derivative of the next question but i want to make sure its right before I move on.

    ii) 2x \ln(3x+1)

    Let u = 2x and v = \ln(3x+1)

    \frac{du}{dx} = 2 and \frac{dv}{dx} = \frac{3}{3x+1}

    F of F

    2 \ln(3x+1) + 2x(\frac{3}{3x+1})

    = 2\ln(3x+1)+2x
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  3. #3
    Member mybrohshi5's Avatar
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    For the second derivative i believe there should be a -x where the -1 is in the numerator and therefore your answer would be -0.345588
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  4. #4
    Member mybrohshi5's Avatar
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    The first derivative of the second one should be.


     2 ln(3x+1) + \frac{6x}{3x+1}


    Use the chain rule and the rule for the derivative of ln(u) which is  \frac {u'}{u}
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  5. #5
    Member mybrohshi5's Avatar
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    If you want me to show you the steps just let me know
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  6. #6
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    Quote Originally Posted by mybrohshi5 View Post
    For the second derivative i believe there should be a -x where the -1 is in the numerator and therefore your answer would be -0.345588
    Can you see where I have gone wrong with this?


    Quote Originally Posted by mybrohshi5 View Post
    The first derivative of the second one should be.



    Use the chain rule and the rule for the derivative of ln(u) which is
    I got that far but obviously tried to simplify it....incorrectly

    Thanks for the input
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  7. #7
    Member mybrohshi5's Avatar
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    Quote Originally Posted by anothernewbie View Post
    Hi would somebody be able to check my answers to some questions I have, i'll just post one at a time as they take up a bit of room:

    Q: Obtain f''(0.3) for each of the following functions.

    i) f(x) = \arccos x

    A:

    f(x) = \arccos x
    f'(x) = -(1-x^2)^ \frac{-1}{2}

    Sub z = 1-x^2
    y = -z^\frac{-1}{2}

    \frac{dy}{dz} = \frac{1}{2}(1-x^2)^\frac{-3}{2}

    right here is where you went wrong. you need to apply the chain rule so you need to include this \frac{dy}{dz} = \frac{1}{2}(1-x^2)^\frac{-3}{2} multiplied by a -2x because the -2x is the derivative of 1-x^2

    The 2's cancel out and you are left with -x on the top.

    f''(x) = \frac{-1}{\sqrt{(1-x^2)^3}}

    f''(0.3) = 1.152
    See above
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  8. #8
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    Aaaah I see I missed a bit whilst copying my working down, thanks for that
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  9. #9
    Member mybrohshi5's Avatar
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    No problem =) Good luck
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