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Thread: logarithmic differentiation

  1. #1
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    logarithmic differentiation

    how do you take the derivative of $\displaystyle y=((x^2+7)/(x^2+8))^{1/7} $?

    I used logarithmic differentiation and got: $\displaystyle (((2x)/(7x^2+49))-((2x)/(7x^2+56)))((x^2+7)/(x^2+8))^{1/7} $

    Doesn't seem to be correct, though..
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  2. #2
    Senior Member apcalculus's Avatar
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    Have you tried checking your work in WolframAlpha?

    http://www.wolframalpha.com/input/?i=differentiate+y%3D((x^2%2B7)/(x^2%2B8))^{1/7

    Good luck!
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  3. #3
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    Quote Originally Posted by cdlegendary View Post
    how do you take the derivative of $\displaystyle y=((x^2+7)/(x^2+8))^{1/7} $?

    I used logarithmic differentiation and got: $\displaystyle (((2x)/(7x^2+49))-((2x)/(7x^2+56)))((x^2+7)/(x^2+8))^{1/7} $

    Doesn't seem to be correct, though..
    $\displaystyle ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)$
    so $\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}$.

    I cannot see any reason to have the "7" multiplied in the denominator of the first fraction.
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  4. #4
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    derivative



    i need to take the derivative of the above equation...none of the results seem to be working. and wolfram alpha gets a strange result, too.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    $\displaystyle ln(y)= ln(x^2+ 7)- (1/7)ln(x^2+ 8)$
    so $\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{2x}{x^2+7}- \frac{2x}{7(x^2+ 8)}$.

    I cannot see any reason to have the "7" multiplied in the denominator of the first fraction.
    Actually since

    $\displaystyle y = \left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}$

    That means

    $\displaystyle \ln{y} = \ln{\left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}}$

    $\displaystyle = \frac{1}{7}\ln{\left(\frac{x^2 + 7}{x^2 + 8}\right)}$

    $\displaystyle = \frac{1}{7}[\ln{(x^2 + 7)} - \ln{(x^2 + 8)}]$

    $\displaystyle = \frac{1}{7}\ln{(x^2 + 7)} - \frac{1}{7}\ln{(x^2 + 8)}$.


    Therefore

    $\displaystyle \frac{1}{y}\,\frac{dy}{dx} = \frac{2x}{7(x^2 + 7)} - \frac{2x}{7(x^2 + 8)}$

    $\displaystyle = \frac{2x(x^2 + 8) - 2x(x^2 + 7)}{7(x^2 + 7)(x^2 + 8)}$

    $\displaystyle = \frac{2x(x^2 + 8 - x^2 - 7)}{7(x^2 + 7)(x^2 + 8)}$

    $\displaystyle = \frac{2x}{7(x^2 + 7)(x^2 + 8)}$.


    Therefore

    $\displaystyle \frac{dy}{dx} = \frac{2xy}{7(x^2 + 7)(x^2 + 8)}$

    $\displaystyle = \frac{2x\left(\frac{x^2 + 7}{x^2 + 8}\right)^{\frac{1}{7}}}{7(x^2 + 7)(x^2 + 8)}$

    $\displaystyle = \frac{\frac{2x(x^2+7)^{\frac{1}{7}}}{(x^2 + 8)^{\frac{1}{7}}}}{7(x^2 + 7)(x^2 + 8)}$

    $\displaystyle = \frac{2x(x^2 + 7)^{\frac{1}{7}}}{7(x^2 + 7)(x^2 + 8)(x^2 + 8)^{\frac{1}{7}}}$

    $\displaystyle = \frac{2x}{7(x^2 + 7)^{\frac{6}{7}}(x^2+ 8)^{\frac{8}{7}}}$.
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