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Math Help - Logarithmic Differentiation: Finding derivative

  1. #1
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    Logarithmic Differentiation: Finding derivative

    y = (x3√(x2 + 1))/(x - 1)
    What I did:
    I used natural log on both sides.
    I brought the exponents into the coefficient.
    I also brought the (x - 1) up by doing it to the negative one exponent
    Then I used the log properties.

    So tell me if this is the correct way:
    ln y = 3lnx + (1/2)ln(x2 + 1) - ln(x - 1)

    Then I find the derivative
    1/y(dy/dx) = 3/x + (1/2)(1/(x2 + 1)) - (1/(x - 1))

    After that I multiplied both sides by y, then plugged in y, which is the original equation.
    So would this be right?

    Thanks
    Last edited by Chaim; March 15th 2013 at 09:41 AM.
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  2. #2
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    Re: Logarithmic Differentiation: Finding derivative

    You made a small mistake in the derivative of ln(x2+1)1/2 but other than that it is correct
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    Re: Logarithmic Differentiation: Finding derivative

    Quote Originally Posted by Shakarri View Post
    You made a small mistake in the derivative of ln(x2+1)1/2 but other than that it is correct
    Hmm, so first I would put the exponent into the coefficient right?
    (1/2)ln(x2+1)

    Um, would I then have the log go into the x and 1?
    Yeah sorry, doing the derivative of 1 variable for ln is easy for me.

    So if it was that (1/2)ln(x2+1)
    I wouldn't just do what's in the parenthesis and put it all down under 1?
    Then multiply it by 1/2?
    So like 1/(2(x2​+1))?
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  4. #4
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    Re: Logarithmic Differentiation: Finding derivative

    v= ln(x^2+1)

    let u=(x^2+1)

    \frac{du}{dx}=2x

    v=ln(u)

    \frac{dv}{du}=\frac{1}{u}

    \frac{dv}{dx}= \frac{du}{dx}\cdot \frac{dv}{du}

    \frac{dv}{dx}=2x \cdot \frac{1}{u}= \frac{2x}{(x^2+1)}
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  5. #5
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    Re: Logarithmic Differentiation: Finding derivative

    Quote Originally Posted by Shakarri View Post
    v= ln(x^2+1)

    let u=(x^2+1)

    \frac{du}{dx}=2x

    v=ln(u)

    \frac{dv}{du}=\frac{1}{u}

    \frac{dv}{dx}= \frac{du}{dx}\cdot \frac{dv}{du}

    \frac{dv}{dx}=2x \cdot \frac{1}{u}= \frac{2x}{(x^2+1)}
    Oh right, now I remember!
    So it's the derivative of that over the original!

    Thanks!


    Then after that you would multiply it by 1/2 right?
    Which will be 2x/(2(x2​+1))
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