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Core Maths 3: Differentiation using the quotient rule

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    Hi there

    Please could someone tell me if the below is correct;



    Many thanks
    Jackie
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    I haven't checked the algebra but how did you go from the fraction at the bottom left to the next stage ? The question doesn't give us x.
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    (Original post by jackie11)
    Hi there

    Please could someone tell me if the below is correct;

    Many thanks
    Jackie
    According to the question you should leave your answer as  \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

    Why did you wrote  \displaystyle \frac{1+1-1}{1+1+1} \ ?
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    (Original post by raheem94)
    According to the question you should leave your answer as  \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

    Why did you sub in x=1?
    He didn't sub in x=1 otherwise the top would be 1 + 2 - 1 = 2

    He could make into a proper fraction I guess.
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    (Original post by member910132)
    He didn't sub in x=1 otherwise the top would be 1 + 2 - 1 = 2

    He could make into a proper fraction I guess.
    Thanks for correcting me, do you understand what he was trying to do?

    This can easily be made into a proper fraction, but i don't think he was trying to do this.
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    (Original post by raheem94)
    According to the question you should leave your answer as  \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

    Why did you sub in x=1?
    ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

    10/10, it becomes 1/1

    So that's what I did for x^2/x^2 and 2x/2x
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    (Original post by jackie11)
    Hi there

    Please could someone tell me if the below is correct;



    Many thanks
    Jackie
    Jackie, Jackie, Jackie

    You CANNOT do that

    You CANNOT cancel the x^2 and then cancel the 2x's ... NO NO NO NO
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    (Original post by jackie11)
    ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

    10/10, it becomes 1/1

    So that's what I did for x^2/x^2 and 2x/2x
    10/10 becomes 1/1 but here the top has -1 and the bottom has +1.

    If you want to simplify it, then you can do it in the following way,
     \displaystyle \frac{x^2+2x-1}{x^2+2x+1}=\frac{x^2+2x+1-1-1}{x^2+2x+1}=\frac{x^2+2x+1}{x^2  +2x+1}-\frac2{x^2+2x+1}= \\  1-\frac2{x^2+2x+1}
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    (Original post by jackie11)
    ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

    10/10, it becomes 1/1

    So that's what I did for x^2/x^2 and 2x/2x
    No you can't do that. say if we have \dfrac{x+3}{x+5}, that ISN'T equal to \dfrac{1+3}{1+5}

    You can only simplify a fraction if the top and bottom are multiplied by the same factor, not 'added on' by the same amount.

    E.g. the above isn't allowed, but this is:

    \dfrac{x(x+3)}{x(x+5)} = \dfrac{1(x+3)}{1(x+5)}

    EDIT: I'm sure you probably do know this, and are just getting confused but hopefully the examples helped
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    aww thank you all so much

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