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    Hi, I did a past paper the AQA C3 June 2012 one and I did everything fine apart from when I got to the final question and I didn't have a clue what to do.

    The questions which I didn't understand are c and all of d.
    Is there a specific name for this type of differentiation with stuff on the bottom, and differentiating tan^-1(ax-b)
    as well as differentiating to find the second derivative when it is a fraction like the dy/dx in C.

    If there isn't a name for the topic does anyone know where I can find some additional resources with more questions which I can practise on? I don't know how to search for these types of questions as its the more advanced differentiation of C3 which I can't do.
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    (Original post by Sayless)
    Hi, I did a past paper the AQA C3 June 2012 one and I did everything fine apart from when I got to the final question and I didn't have a clue what to do.

    The questions which I didn't understand are c and all of d.
    Is there a specific name for this type of differentiation with stuff on the bottom, and differentiating tan^-1(ax-b)
    as well as differentiating to find the second derivative when it is a fraction like the dy/dx in C.

    If there isn't a name for the topic does anyone know where I can find some additional resources with more questions which I can practise on? I don't know how to search for these types of questions as its the more advanced differentiation of C3 which I can't do.
    this is prelude of implicit differentiation which will eventually cover in your board in C4

    In C3 you laern that

    if x = f(y), then dy/dx = 1 / dx/dy
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    (Original post by TeeEm)
    this is prelude of implicit differentiation which will eventually cover in your board in C4

    In C3 you laern that

    if x = f(y), then dy/dx = 1 / dx/dy
    So these questions are both 1/dx/dy questions, still for d(ii) I wouldn't have a clue how to differentiate something like that i've never seen it before
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    (Original post by Sayless)
    So these questions are both 1/dx/dy questions, still for d(ii) I wouldn't have a clue how to differentiate something like that i've never seen it before
    For d)ii)
    You simply need to differentiate:
    \dfrac{1}{1+(x-1)^2}-\dfrac{d}{dx} \left(\dfrac{d}{dx}(\ln(x)) \right)

    As for the derivative of \arctan(x), you should calculate the derivative at least once using the method TeeEm suggested, but from then on, I would consider it a quotable result, and I imagine it's in your formula book.
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    (Original post by joostan)
    For d)ii)
    You simply need to differentiate:
    \dfrac{1}{1+(x-1)^2}-\dfrac{d}{dx} \left(\dfrac{d}{dx}(\ln(x)) \right)

    As for the derivative of \arctan(x), you should calculate the derivative at least once using the method TeeEm suggested, but from then on, I would consider it a quotable result, and I imagine it's in your formula book.
    Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
    I can't do TeeEm's method on y=tan^-1(x-1)
    I ended up with dy/dx = 1/sec^2 y
    when I use it from the formula booklet I get the correct answer though
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    For part di) I got x=2 and x=1 for part dii) do I need to use quotient rule
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    (Original post by Sayless)
    Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
    I can't do TeeEm's method on y=tan^-1(x-1)
    I ended up with dy/dx = 1/sec^2 y
    when I use it from the formula booklet I get the correct answer though
    Yes and you found that sec^2y is x^2-2x+2 so dy/dx=1/x^2-2x+2
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    (Original post by Sayless)
    Yeah, I have to differentiate that, but I don't know how to, I can differentiate using chain rule, product rule, quotient rule, but for stuff with stuff squared underneath in a fraction and another fraction I haven't learnt how to differentiate that, do you have a guide or something I could use.
    I can't do TeeEm's method on y=tan^-1(x-1)
    I ended up with dy/dx = 1/sec^2 y
    when I use it from the formula booklet I get the correct answer though
    Look in your formula booklet: it should say that \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \tan^{-1}x = \frac{1}{1+x^2} - so in your case, just replace the x with x-1.
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    (Original post by Hassan55)
    Yes and you found that sec^2y is x^2-2x+2 so dy/dx=1/x^2-2x+2
    Ah right yeah I get that part now thanks

    (Original post by Zacken)
    Look in your formula booklet: it should say that \displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \tan^{-1}x = \frac{1}{1+x^2} - so in your case, just replace the x with x-1.
    I managed to do that now, stuck with d(ii) only now
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    Part dii) dy/dx=(1/x^2-2x+2) - (1/x) so differentiating first fraction using quotient rule u=1 v=x^2-2x+2 and second fraction would just be -lnx
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    (Original post by Sayless)
    Ah right yeah I get that part now thanks



    I managed to do that now, stuck with d(ii) only now
    Well, you need to evalute \frac{d}{dx} (\arctan x - \ln x) then once you've got that, differentiate whatever you have again - what do you get?
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    (Original post by Hassan55)
    For part di) I got x=2 and x=1 for part dii) do I need to use quotient rule
    You don't HAVE to use the quotient rule, but you may as well.
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    (Original post by joostan)
    You don't HAVE to use the quotient rule, but you may as well.
    What other methods are there in order to differentiate a quotient?
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    (Original post by Zacken)
    Well, you need to evalute \frac{d}{dx} (\arctan x - \ln x) then once you've got that, differentiate whatever you have again - what do you get?
    I'm trying to do the quotient rule for it, I get
    u=1
    v= x^2 -2x + 2
    du/dx = 0
    dv/dx = 2x-2

    (x^2-2x+2)(0) - (1)(2x-2) / (x^2-2x+2)^2

    = -2x+2 / (x^2-2x+2)^2

    then for the 1/x i get x^-2

    so overall its
    -2x+2/(x^2-2x+2)^2 + x^-2

    is that right
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    (Original post by Sayless)
    I'm trying to do the quotient rule for it, I get
    u=1
    v= x^2 -2x + 2
    du/dx = 0
    dv/dx = 2x-2

    (x^2-2x+2)(0) - (1)(2x-2) / (x^2-2x+2)^2

    = -2x+2 / (x^2-2x+2)^2

    then for the 1/x i get x^-2

    so overall its
    -2x+2/(x^2-2x+2)^2 + x^-2

    is that right
    Almost d/dx(1/x) = -x^(-2) not +
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    (Original post by Hassan55)
    What other methods are there in order to differentiate a quotient?
    This is simply the chain rule with (x^2 - blah blah)^{-1}
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    (Original post by Zacken)
    This is simply the chain rule with (x^2 - blah blah)^{-1}
    Oh yeah of course, thanks
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    (Original post by Zacken)
    Almost d/dx(1/x) = -x^(-2) not +
    Thanks, i got it now
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    (Original post by Sayless)
    Thanks, i got it now
    Great!
 
 
 
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