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    I understand how to do it for say y = (tan(1+x)^-1) but this fraction is really throwing me off... any advice?
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    (Original post by 2014_GCSE)
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    I understand how to do it for say y = (tan(1+x)^-1) but this fraction is really throwing me off... any advice?
    2014 AQA right? Anyway, just approach it as normal, it falls out nicely (From memory at least)
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    write f(x) = tan(y)

    make x the subject, then find dx/dy.... after that you can get dy/dx easily.
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    In the first part you have shown that
     \displaystyle \frac{dy}{dx} = \frac{1}{1+x^2}
    So now just solve the differential equation giving  y= \arctan x + c .
    y is defined above as  \arctan \left  ( \frac{1+x}{1-x} \right ) so substitute that in for y. Then find c as you normally would and rearrange to give the required result.
    ^ for the second part of the question.

    For the first part just use chain rule (you will need quotient rule as well when differentiating the fraction).
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    Thanks, zetamcfc, the bear and B_9710. I managed to get part a) with what you guys said but still struggling with part b). I even just had a look at the mark scheme and I still am not really understanding b)...


    For the post above, where have you pulled the equation "y=arctanx + c" from? I understand it may give you the answer but I don't understand the logical steps?
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    (Original post by 2014_GCSE)
    Thanks, zetamcfc, the bear and B_9710. I managed to get part a) with what you guys said but still struggling with part b). I even just had a look at the mark scheme and I still am not really understanding b)...


    For the post above, where have you pulled the equation "y=arctanx + c" from? I understand it may give you the answer but I don't understand the logical steps?
    What's part b?
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    (Original post by Kvothe the arcane)
    What's part b?
    Sorry, I didn't realise the photo didn't have the letters in there. Part b) is "Hence, given that x<1... etc."
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    (Original post by 2014_GCSE)
    Sorry, I didn't realise the photo didn't have the letters in there. Part b) is "Hence, given that x<1... etc."
    We need to know what ...etc is

    Edit: stupid me :facepalm2:
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    (Original post by 2014_GCSE)
    Sorry, I didn't realise the photo didn't have the letters in there. Part b) is "Hence, given that x<1... etc."
    Integrate your dy/dx, what do you see?
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    (Original post by 2014_GCSE)
    Thanks, zetamcfc, the bear and B_9710. I managed to get part a) with what you guys said but still struggling with part b). I even just had a look at the mark scheme and I still am not really understanding b)...


    For the post above, where have you pulled the equation "y=arctanx + c" from? I understand it may give you the answer but I don't understand the logical steps?
    It comes from solving the differential equation  \displaystyle y=\frac{1}{1+x^2} .
    And remember that  \displaystyle \int \frac{1}{1+x^2} dx = \arctan x +c .
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    Ah, of course. Got it!! Thanks everybody

    (Original post by Kvothe the arcane)
    We need to know what ...etc is
    I put the question in the OP
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    (Original post by 2014_GCSE)
    Ah, of course. Got it!! Thanks everybody

    I put the question in the OP
    I wouldn't do it the way that the other users have, I'd write f(x) = \arctan \frac{1+x}{1-x} - \arctan x, then from the first part, notice that:

    \displaystyle f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\arctan \frac{1+x}{1-x} - \arctan x \right) = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0

    So that f'(x) = 0 \Rightarrow f(x) = c. Then f(0) = \arctan{1} - \arctan{0} = \frac{\pi}{4}.
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    (Original post by Zacken)
    I wouldn't do it the way that the other users have, I'd write f(x) = \arctan \frac{1+x}{1-x} - \arctan x, then from the first part, notice that:

    \displaystyle f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\arctan \frac{1+x}{1-x} - \arctan x \right) = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0

    So that f'(x) = 0 \Rightarrow f(x) = c. Then f(0) = \arctan{1} - \arctan{0} = \frac{\pi}{4}.
    STEP I question there?
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    (Original post by Zacken)
    I wouldn't do it the way that the other users have, I'd write f(x) = \arctan \frac{1+x}{1-x} - \arctan x, then from the first part, notice that:

    \displaystyle f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\arctan \frac{1+x}{1-x} - \arctan x \right) = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0

    So that f'(x) = 0 \Rightarrow f(x) = c. Then f(0) = \arctan{1} - \arctan{0} = \frac{\pi}{4}.
    Why would you know that f'(x)=0?

    I assume you're not used that fact f(x)=\frac{\pi}{4} as that'd be circular reasoning.
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    (Original post by Kvothe the arcane)
    Why would you know that f'(x)=0?

    I assume you're not used that fact f(x)=\dfrac{\pi}{4} as that'd be circular reasoning.
    Because (\arctan x)' = \frac{1}{1+x^2} and \left(\arctan \frac{1+x}{1-x}\right)' = \frac{1}{1+x^2}.

    Derivative is a linear operator so since you're differentiating the difference of two things whose derivatives are the same, then f'(x) = 0. So f(x) is a constant and plugging in any value of x will reveal the value of this constant.
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    (Original post by B_9710)
    STEP I question there?
    First part of a STEP I question from a paper in the 80's or 90's iirc.
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    (Original post by Zacken)
    Because (\arctan x)' = \frac{1}{1+x^2} and \left(\arctan \frac{1+x}{1-x}\right)' = \frac{1}{1+x^2}.

    Derivative is a linear operator so since you're differentiating the difference of two things whose derivatives are the same, then f'(x) = 0. So f(x) is a constant and plugging in any value of x will reveal the value of this constant.
    That was stupid of me. Didn't notice they were the same. Thanks. Noted.
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    (Original post by Kvothe the arcane)
    That was stupid of me. Didn't notice they were the same. Thanks. Noted.
    I think that was the intended solution as well - they gave you part (a) to show the derivative of that is the derivative of \arctan x too. It's a nice concept.
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    (Original post by Zacken)
    I think that was the intended solution as well - they gave you part (a) to show the derivative of that is the derivative of \arctan x too. It's a nice concept.
    http://filestore.aqa.org.uk/subjects...W-MS-JUN14.PDF

    Q7 if you want the desired solution
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    (Original post by zetamcfc)
    http://filestore.aqa.org.uk/subjects...W-MS-JUN14.PDF

    Q7 if you want the desired solution
    Thanks for setting me straight. :lol:
 
 
 
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