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    how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
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    (Original post by hajs)
    how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
    You can ignore the modulus within the ln when differentiating. It is introduced during integration.
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    (Original post by hajs)
    how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
    x^2 + 1 is always positive, so don't worry too much about the modulus sign.

    set, u = x^2 + 1, then y = ln u.

    Plug into the chain rule.
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    (Original post by hajs)
    how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
    Differentiate lnu where u = x^2 + 1

    Becomes [1/u]*(du/dx) iirc
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    (Original post by morgan8002)
    You can ignore the modulus within the ln when differentiating. It is introduced during integration.
    no way! I did not know that! haha okay so that would be 2x right?? so why does the book say y = ln|x^2 +1| dy/dx = 1/x^2+1 * 2x ??
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    nevermind this......................
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    (Original post by hajs)
    how do you differentiate ln | x^2 + 1| ?? i know its the chain rule but how??
    If x is real then |x^2 +1| = x^2 + 1 so you can just ignore the modulus

    Just apply the standard chain rule with y = ln u where u = x^2 + 1.
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    (Original post by Mr Inquisitive)
    Differentiate lnu where u = x^2 + 1

    Becomes [1/(du/dx)]*u iirc
    Actually it's \frac{du}{dx}\frac{1}{u}
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    (Original post by Lolgarithms)
    x^2 + 1 is always positive, so don't worry too much about the modulus sign.

    set, u = x^2 + 1, then y = ln u.

    Plug into the chain rule.
    OHHHHH. oh yeah my bad haha. thank you!!!!
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     \displaystyle \frac{d}{dx} (ln(f(x))= \frac{f ' (x)}{f(x)}

    the "mod" doesn`t matter: x^2+1 is always +ve..
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    (Original post by morgan8002)
    Actually it's \frac{du}{dx}\frac{1}{u}
    That's the one... Been two years
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    (Original post by morgan8002)
    Actually it's \frac{du}{dx}\frac{1}{u}
    (Original post by davros)
    If x is real then |x^2 +1| = x^2 + 1 so you can just ignore the modulus

    Just apply the standard chain rule with y = ln u where u = x^2 + 1.
    thanks i get it!!
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    (Original post by hajs)
    no way! I did not know that! haha okay so that would be 2x right?? so why does the book say y = ln|x^2 +1| dy/dx = 1/x^2+1 * 2x ??
    You can ignore the modulus, but use the substitution u = x^2 + 1.
 
 
 
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