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Math Help - Help Natural Logs

  1. #1
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    Help Natural Logs

    Hi, please help

    solve the equation f'(x) = 0

    f(x)= (ln x +2x) ^ 1/3


    thanks
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  2. #2
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    Quote Originally Posted by thelazyman
    Hi, please help

    solve the equation f'(x) = 0

    f(x)= (ln x +2x) ^ 1/3


    thanks
    Given the function,
    f(x)=(\ln x+2x)^{1/3}
    Then, by the Generalized Exponent Rule,
    f'(x)=(1/3)(1/x+2)(\ln x+2x)^{-2/3}
    But f'(x)=0
    Thus,
    (1/3)(1/x+2)(\ln x+2x)^{-2/3}=0
    thus,
    (1/x+2)=0
    because (\ln x+2x)^{-2/3} is positive.
    Thus,
    x=-1/2

    This is my 2 Post!
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  3. #3
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    Natural Logs again

    Hi, test is tommorow need help with this question

    For each of the following, find the slope of the tangent.

    For the curve defined by xe^y + ylnx = 2 at (1, ln2).

    Thanks
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  4. #4
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    Hi, I understand how you got the -1/2, but what happens to the 1/3 and (lnx + 2x)^ -2/3, do they cancel out and why>?
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  5. #5
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    Quote Originally Posted by thelazyman
    Hi, I understand how you got the -1/2, but what happens to the 1/3 and (lnx + 2x)^ -2/3, do they cancel out and why>?
    You could think of it like that.
    See the problem is that (\ln x+2x)^{-2/3} is never zero thus you can divide both sides by this.
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  6. #6
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    Quote Originally Posted by thelazyman
    Hi, test is tommorow need help with this question

    For each of the following, find the slope of the tangent.

    For the curve defined by xe^y + ylnx = 2 at (1, ln2).

    Thanks
    Take the derivative of both sides implicitly,
    Remember to use the product rule
    \frac{d}{dx}(x)e^y+xe^y\frac{dy}{dx}+\frac{d}{dx}(  \ln x)y+\ln x\frac{dy}{dx}=\frac{d}{dx}(2)
    Simplify,
    e^y+xe^y\frac{dy}{dx}+\frac{1}{x}y+\ln x\frac{dy}{dx}=0
    Reorganize,
    xe^y\frac{dy}{dx}+\ln x\frac{dy}{dx}=-e^y-\frac{y}{x}
    Now factor,
    \frac{dy}{dx}(xe^y+\ln x)=-e^y-\frac{y}{x}
    Remember the problem says that point (x,y)=(1,\ln 2)
    Substitute those values for x,y
    Thus, (remember e^{\ln 2}=2,\ln 1=0)
    \frac{dy}{dx}((1)(2)+0)=-2-\ln 2
    Thus,
    \frac{dy}{dx}=\frac{-2-\ln 2}{2}
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