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Math Help - derivative of a logarithmic function

  1. #1
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    derivative of a logarithmic function

    How is it done?
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  2. #2
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    Re: derivative of a logarithmic function

    I have done it. It is done the usual way, derivative of (x + (x^2+k)^0.5)/(x + (x^2+k)^0.5).
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  3. #3
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    Re: derivative of a logarithmic function

    Use:

    \frac{d}{dx}\left(\ln(f(x))\right)=\frac{1}{f(x)} \cdot\frac{df}{dx}
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  4. #4
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    Re: derivative of a logarithmic function

    Hello, Stuck Man!

    \text{Show that the derivative of }\,y \:=\:\ln|x+\sqrt{x^2+k}| \,\text{ is: }\<y' \:=\:\frac{1}{\sqrt{x^2+k}}

    First, you do the Calculus (find the derivative),
    . . then do some Algebra.


    We have: . y \:=\:\ln\left|x + (x^2+k)^{\frac{1}{2}}\right|


    The derivative is: . y' \:=\:\frac{1+\frac{1}{2}(x^2+k)^{-\frac{1}{2}}\cdot2x}{x+(x^2+k)^{\frac{1}{2}}} \;=\;\dfrac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x+\sqrt{x^2+k}}

    Multiply by \frac{\sqrt{x^2+k}}{\sqrt{x^2+k}}:\;\;\frac{\sqrt{  x^2+k}}{\sqrt{x^2+k}}\cdot\frac{1 + \dfrac{x}{\sqrt{x^2+k}}}{x + \sqrt{x^2+k}}

    . . . . . . . . . . . =\;\frac{\sqrt{x^2+k}+x}{\sqrt{x^2+k}\,(x + \sqrt{x^2+k})}

    . . And cancel: . \frac{1}{\sqrt{x^2+k}}

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