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Math Help - derivative problem with ln

  1. #1
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    derivative problem with ln

    Differentiate. .



    1/(xsqrt(x^2-2))(sqrt(x^2-2) + (2x^2)/2sqrt(x^2-2)

    common denominator: 2sqrt(x^2-2)
    = 1/(xsqrt(x^2-2)) {[2(x^2-2)+x^2 ]/[2sqrt(x^2-2)}

    getting rid of the other denominator
    = 3x^2-4/(2*sqrt(x^2-2))*(1/1/(xsqrt(x^2-2))
    = 3x^2-4/ 2x(x^2-2)
    = 3x^2-4/2x^3-4x
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  2. #2
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    Quote Originally Posted by hazecraze View Post
    Differentiate. .



    1/(xsqrt(x^2-2))(sqrt(x^2-2) + (2x^2)/2sqrt(x^2-2)

    common denominator: 2sqrt(x^2-2)
    = 1/(xsqrt(x^2-2)) {[2(x^2-2)+x^2 ]/[2sqrt(x^2-2)}

    You seem to have an error in here when you calculated the derivative of \sqrt{x^2-2}. It seems like the 2 should have been canceled.

    getting rid of the other denominator
    = 3x^2-4/(2*sqrt(x^2-2))*(1/1/(xsqrt(x^2-2))
    = 3x^2-4/ 2x(x^2-2)
    = 3x^2-4/2x^3-4x
    g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+x\frac{d}{dx}\sqrt{x^2-2})

    \frac{d}{dx}\sqrt{x^2-2}=\frac{2x}{2\sqrt{x^2-2}}=\frac{x}{\sqrt{x^2-2}}

    So this becomes:

    g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+\frac{x^2}{\sqrt{x^2-2}})
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+x\frac{d}{dx}\sqrt{x^2-2})

    \frac{d}{dx}\sqrt{x^2-2}=\frac{2x}{2\sqrt{x^2-2}}=\frac{x}{\sqrt{x^2-2}}

    So this becomes:

    g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+\frac{x^2}{\sqrt{x^2-2}})
    Ohh, so... g'(x)=\frac{2x^2-2}{2\sqrt{x^2-2}}
    cancel out the 2s

    g'(x)=\frac{x^2-1}{\sqrt{x^2-2}}

    =

    =\frac{x^2-1}{\sqrt{x^2-2}}* =\frac{1}{x\sqrt{x^2-2}}

    = g'(x)=\frac{x^2-1}{x^3-2}
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  4. #4
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    Quote Originally Posted by hazecraze View Post
    Ohh, so... g'(x)=\frac{2x^2-2}{2\sqrt{x^2-2}}
    cancel out the 2s

    g'(x)=\frac{x^2-1}{\sqrt{x^2-2}}

    =

    =\frac{x^2-1}{\sqrt{x^2-2}}* =\frac{1}{x\sqrt{x^2-2}}

    = g'(x)=\frac{x^2-1}{x^3-2}
    You simplified this wrong:



    Adding inside the parenthesis:

    g'(x)=\frac{1}{x\sqrt{x^2+2}}(\frac{2x^2-2}{\sqrt{x^2-2}})

    Now just multiply the fractions:

    (this is where you messed up I think. You added the denominators instead of multiplying.)

    g'(x)=\frac{2x^2-2}{x(x^2-2)}
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