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    Compute the derivative \frac{d}{dx}\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]

    Let y=\frac{x^2+1}{x^3(x-1)^2}

    Then using logarithmic differentiation let \ln y=\ln\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]=\ln(x^2+1)-3\ln x-2\ln(x-1)

    \frac{d}{dx}(\ln y) = \frac{d}{dx}\big[\ln(x^2+1)-3\ln x-2\ln(x-1)\big]

    \frac{1}{y}\frac{dy}{dx} = \frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}

    \frac{dy}{dx}=y\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big]

    which equals \frac{x^2+1}{x^3(x-1)}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big] ? I'm a bit lost now, did I miss something? Wolfram is giving me a different answer...
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    (Original post by AishaGirl)
    Compute the derivative \frac{d}{dx}\Big[\frac{x^2+1}{x^3(x-1)^2}\Big]
    The expressions will be equivalent, you're answer is correct.
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    (Original post by B_9710)
    The expressions will be equivalent, you're answer is correct.
    My answer? I've not even finished differentiating yet. I must be missing something because when I try to get rid of the square brackets it's wrong.
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    (Original post by AishaGirl)
    My answer? I've not even finished differentiating yet. I must be missing something because when I try to get rid of the square brackets it's wrong.
    You have finished the differentiation part - next it's just algebra. It's not wrong, I checked your answer and it's correct, it's must just be in a different form to the 'correct' answer that you saw.
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    (Original post by B_9710)
    You have finished the differentiation part - next it's just algebra. It's not wrong, I checked your answer and it's correct, it's must just be in a different form to the 'correct' answer that you saw.
    Wolfram alpha says that it's \frac{2}{x^3(x-1)^2} for \frac{x^2+1}{x^3(x-1)} where is this coming from? Is it x^3(x-1)^2 or x^3(x-1)?

    I'm so confused. I've missed something and I think you have too?
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    Oh I have!

    It's
    \frac{x^2+1}{x^3(x-1)^2}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big] which gives \frac{2}{x^2(x-1)^2}-\frac{3(x^2+1}{x^4(x-1)^2}-\frac{2(x^2+1)}{x^3(x-1)^3}

    =\frac{2x^2(x-1)-3(x^2+1)(x-1)-2(x^2+1)x}{x^3(x-1)^3}

    and finally \frac{-3x^3+x^2-5x+3}{x^4(x-1)^3}

    which now matches what wolfram alpha says.

    I was missing a power.
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    (Original post by AishaGirl)
    Oh I have!

    It's
    \frac{x^2+1}{x^3(x-1)^2}\Big[\frac{2x}{x^2+1}-\frac{3}{x}-\frac{2}{x-1}\Big] which gives \frac{2}{x^2(x-1)^2}-\frac{3(x^2+1}{x^4(x-1)^2}-\frac{2(x^2+1)}{x^3(x-1)^3}

    =\frac{2x^2(x-1)-3(x^2+1)(x-1)-2(x^2+1)x}{x^3(x-1)^3}

    and finally \frac{-3x^3+x^2-5x+3}{x^4(x-1)^3}

    which now matches what wolfram alpha says.

    I was missing a power.
    Right. I did notice you forgot the power of 2 of that term but I forgot to mention it - it was of course all right apart from that as you said.
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    Is logarithmic differentiation useful in A level? Or is this uni?
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    (Original post by thekidwhogames)
    Is logarithmic differentiation useful in A level? Or is this uni?
    it's useful for A2 Maths in C3, C4 and DE for MEI maths, dk about other boards though
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    (Original post by thekidwhogames)
    Is logarithmic differentiation useful in A level? Or is this uni?
    no need for the bump but yeah applying log to both sides can be useful at A-level to avoid getting lost in a mess of algebra
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    (Original post by _gcx)
    no need for the bump but yeah applying log to both sides can be useful at A-level to avoid getting lost in a mess of algebra
    Alright, thanks!
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    (Original post by NishatM)
    it's useful for A2 Maths in C3, C4 and DE for MEI maths, dk about other boards though
    Yeah sorry about that. I was watching a video by Prof Leonard and he went over this so it seemed to be really useful. Thanks for the help!
 
 
 
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