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Math Help - Derivative of Logarithmic Function

  1. #1
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    Derivative of Logarithmic Function

    Differentiate h(x) = \ln{(x + \sqrt{x^2-6})}

    My work:

    let u = x + \sqrt{x^2-6} then y = \ln{u}

    \displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx  } = \frac{1}{x + \sqrt{x^2-6}}(1+\frac{1}{2}(x^2-6)^{-1/2}(2x))

    \displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}

    \displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}

    \displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}

    Where did I go wrong??? Thanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by joatmon View Post
    Differentiate h(x) = \ln{(x + \sqrt{x^2-6})}

    My work:

    let u = x + \sqrt{x^2-6} then y = \ln{u}

    \displaystyle\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\frac{1}{u}\frac{du}{dx  } = \frac{1}{x + \sqrt{x^2-6}}(\boxed{1}+\frac{1}{2}(x^2-6)^{-1/2}(2x))

    \displaystyle=\frac{2x}{(x+\sqrt{x^2-6})(2)\sqrt{x^2-6}}

    \displaystyle=\frac{x}{(x+\sqrt{x^2-6})\sqrt{x^2-6}}

    \displaystyle=\frac{x}{(x\sqrt{x^2-6} + x^2-6}

    Where did I go wrong??? Thanks.
    You forgot about the term I boxed in your work.

    The final answer should be \dfrac{1}{x + \sqrt{x^2-6}}+\dfrac{x}{x\sqrt{x^2-6} + x^2-6}=\dfrac{x+\sqrt{x^2-6}}{x\sqrt{x^2-6}+x^2-6}=\dfrac{1}{\sqrt{x^2-6}}

    I hope this makes sense.
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