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Math Help - Differentiate with respect to x

  1. #1
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    Differentiate with respect to x

    Please could someone show me how to
    Differentiate with respect to x

    Y = Ln (x^2+1)/(x^2-1)

    Thanks in advance,

    HNC



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  2. #2
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    Quote Originally Posted by HNCMATHS View Post
    Please could someone show me how to
    Differentiate with respect to x

    Y = Ln (x^2+1)/(x^2-1)

    Thanks in advance,

    HNC



    Quotient Rule
    Last edited by mr fantastic; January 28th 2011 at 07:04 PM.
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  3. #3
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    Quote Originally Posted by HNCMATHS View Post
    Please could someone show me how to
    Differentiate with respect to x

    Y = Ln (x^2+1)/(x^2-1)

    Thanks in advance,

    HNC



    y=ln\left(\frac{x^2+1}{x^2-1}\right)

    or

    y=\frac{ln(x^2+1)}{x^2-1}\;\;?
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  4. #4
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    y=ln\left\frac{x^2+1}{x^2-1}\right

    Thanks, sorry for the confusion
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  5. #5
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    Quote Originally Posted by HNCMATHS View Post
    y=ln\left\frac{x^2+1}{x^2-1}\right

    Thanks, sorry for the confusion
    \displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)

    Use the log rules for differentiation.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)

    Use the log rules for differentiation.
    Do you know where I could find them?
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  7. #7
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    \displaystyle\frac{d}{dx}\left[\ln{x}\right]=\frac{\frac{d}{dx}\left[x\right]}{x}
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  8. #8
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    If you are talking about the identity given by dwsmith, you could find it in your precalc book or look online for properties of logarithms

    \ln\bigg(\dfrac{A}{B}\bigg)\;=\;\ln(A)-\ln(B)
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  9. #9
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    Quote Originally Posted by HNCMATHS View Post
    y=ln\left\frac{x^2+1}{x^2-1}\right

    Thanks, sorry for the confusion
    \displaystyle\ u=\frac{x^2+1}{x^2-1}

    y=ln(u)


    Using the chain rule

    \displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du  }{dx}=\frac{1}{u}\frac{du}{dx}


    Using the Quotient Rule

    \displaystyle\frac{du}{dx}=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2-1\right)^2}

    =\displaystyle\frac{2x\left(x^2-x^2-1-1\right)}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2-1\right)^2}


    \displaystyle\frac{dy}{dx}=-\frac{x^2-1}{x^2+1}\;\frac{4x}{\left(x^2-1\right)^2}=-\frac{4x}{\left(x^2+1\right)\left(x^2-1\right)}

    =\displaystyle\frac{4x}{1-x^4}


    Using log laws

    log\frac{a}{b}=loga-logb

    log(ab)=loga+logb

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


    Using the chain rule

    \displaystyle\frac{dy}{dx}=\frac{1}{x^2+1}(2x)-\frac{1}{x^2-1}(2x)=\frac{\left(x^2-1\right)2x-\left(x^2+1\right)2x}{\left(x^2+1\right)\left(x^2-1\right)}

    =\displaystyle\frac{2x^3-2x-2x^3-2x}{x^4-1}
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\ln\left(\frac{x^2+1}{x^2-1}\right)=\ln(x^2+1)-\ln(x^2-1)

    Use the log rules for differentiation.
    Adding to this: The derivative of each log term on the right hand side is easily found using the chain rule.
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  11. #11
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    \displaystyle \left[\ln{f(x)}\right]' = \frac{f'(x)}{f(x)}.
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  12. #12
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    Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still
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  13. #13
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    Quote Originally Posted by HNCMATHS View Post
    Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still
    Have you read any of the previous posts?
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  14. #14
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    Quote Originally Posted by HNCMATHS View Post
    Sorry guys, really dont understand this one... I want to use the laws of logs to solve it but i'm not sure how to still
    Use the laws of logs to avoid having to mess with fractions.

    You will still need to apply the Chain Rule

    and use

    \frac{d}{dx}ln(x)=\frac{1}{x}

    Then

    u=x^2+1\Rightarrow\frac{d}{dx}ln(u)=\frac{d}{du}ln  (u)\;\frac{du}{dx}

    so

    \frac{d}{du}ln(u)=\frac{1}{u}

    and so on......


    Here's how to use the Chain Rule:

    You've gotten to

    ln\left(x^2+1\right)-ln\left(x^2-1\right)

    The derivative "with respect to x" of ln(x) is \frac{1}{x}

    \frac{d}{dx}ln(x)=\frac{1}{x}

    We want to differentiate ln\left(x^2+1\right) "with respect to x" however.

    If we write u=x^2+1 then we can say we are calculating the derivative of ln(u) "with respect to x".

    However

    \frac{d}{du}ln(u)=\frac{1}{u}

    \frac{d}{dx}ln(x)=\frac{1}{x}

    \frac{d}{dy}ln(y)=\frac{1}{y}

    But what about \frac{d}{dx}ln(u)\;\;?

    We have to use the Chain Rule for that...

    \frac{d}{dx}ln(u)=\frac{d}{du}ln(u)\;\frac{du}{dx}

    We multiply those 2 derivatives and we can calculate both of them.

    \frac{du}{dx}=\frac{d}{dx}\left(x^2+1\right)=2x

    \frac{d}{du}ln(u)=\frac{1}{u}

    Now we replace "u" with it's original expression in x.

    \frac{1}{u}=\frac{1}{x^2+1}

    Therefore

    \frac{d}{dx}ln\left(x^2+1\right)=\frac{2x}{x^2+1} by multiplying the derivatives.

    The same procedure applies to the 2nd log expression. Then the 2 results need to be subtracted.
    Last edited by Archie Meade; January 29th 2011 at 04:56 PM.
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  15. #15
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    Hi Everyone,

    I have done as Archie said, and ended up with...

    =\frac{2x}{x^2+1} - \frac{2x}{x^2-1}

    Now how can I simplify that further? I'm not sure how to subtract one from the other as I quite obviouslly need to,

    Thanks Again,
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