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    (Original post by RDKGames)
    Isn't differentiation from first principles from FP1? Or did I miss somewhere that you'd doing FM? I don't think it's in C1.
    The wjec spec has differentiation in c1
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    (Original post by ihatePE)
    The wjec spec has differentiation in c1
    Well I'm sure every C1 module has differentiation. Normal differentiation and differentiation from first principles are two slightly different things; for one you're dealing with limits and the other you're just moving the exponents down. I'm just surprised C1 has the first principles method, not that it's difficult anyway. Just asking because with AQA first principles and normal were split between C1 and FP1.
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    (Original post by RDKGames)
    Well I'm sure every C1 module has differentiation. Normal differentiation and differentiation from first principles are two slightly different things; for one you're dealing with limits and the other you're just moving the exponents down. I'm just surprised C1 has the first principles method, not that it's difficult anyway. Just asking because with AQA first principles and normal were split between C1 and FP1.
    Yeah i was quite pleasantly surprised when i did the normal differentiation until my teacher pulled out from first principles :hoppy: but its nice to get something difficult out first in c1. Then theres also something about derivative and f''(x)...
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    How does f(x+h) - f(x) / (x+h) - x becomes f(x+h) -f(x) / h ?

    i dont see how the denominator cancels out with anything
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    (Original post by ihatePE)
    How does f(x+h) - f(x) / (x+h) - x becomes f(x+h) -f(x) / h ?

    i dont see how the denominator cancels out with anything
    The x's cancel out.

    (x+h)-x = x-x+h
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    (Original post by RDKGames)
    The x's cancel out.

    (x+h)-x = x-x+h
    ooooh i thought the -x outside was meant to be multiplied with the bracket. oh.
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    (Original post by ihatePE)
    ooooh i thought the -x outside was meant to be multiplied with the bracket. oh.
    Nope. Besides, why would it be multiplied? You are finding the gradient between two points; you don't multiply anything on this stage.
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    (Original post by RDKGames)
    Nope. Besides, why would it be multiplied? You are finding the gradient between two points; you don't multiply anything on this stage.
    i wasnt going to multiply it anyways, i just thought how can u cancel it if the signs are multiplication which is in fact minus

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    can u confirm for me if the first paragraph ''the derivative...'' has any other topics apart from differentiation from first principles?
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    (Original post by ihatePE)
    i wasnt going to multiply it anyways, i just thought how can u cancel it if the signs are multiplication which is in fact minus

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    can u confirm for me if the first paragraph ''the derivative...'' has any other topics apart from differentiation from first principles?
    Just gonna pretend I understood what you said.

    And yes, the rates of changes as well as the second order differential.
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    (Original post by RDKGames)
    Just gonna pretend I understood what you said.

    And yes, the rates of changes as well as the second order differential.
    i know it's so hard to understand and explain what goes on through my head

    and im gonna pretend i understand what you said
    Spoiler:
    Show
    dw i'll get it when i come across it soon :hoppy:
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    (Original post by ihatePE)
    i know it's so hard to understand and explain what goes on through my head

    and im gonna pretend i understand what you said
    Spoiler:
    Show
    dw i'll get it when i come across it soon :hoppy:
    How was I unclear for you to pretend? :facepalm:

    I answered your question as in YES there are topics mentioned within the first paragraph apart from differentiation from first principles; they are rates of change and second differentials. Confirmed, as you asked.
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    (Original post by RDKGames)
    How was I unclear for you to pretend? :facepalm:

    I answered your question as in YES there are topics mentioned within the first paragraph apart from differentiation from first principles; they are rates of change and second differentials. Confirmed, as you asked.
    i was directing it at the topics itself oops but yes u answered my question
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    HAIIIIII

    so i just wanted to know why the second order derivative is d^2y/dx^2 why is the d and x to the power of 2
    actually, what is d to begin with. ive done differientation but dont get why d is in the equation
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    (Original post by ihatePE)
    HAIIIIII

    so i just wanted to know why the second order derivative is d^2y/dx^2 why is the d and x to the power of 2
    actually, what is d to begin with. ive done differientation but dont get why d is in the equation
    d represents an infinitely small change. So dy means an infinitely small change in y and dx is an infinitely small change in x. As you can see, change in y over change in x gives you the gradient... or the rate of change, rather.
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    (Original post by ihatePE)
    HAIIIIII

    so i just wanted to know why the second order derivative is d^2y/dx^2 why is the d and x to the power of 2
    actually, what is d to begin with. ive done differientation but dont get why d is in the equation
    To answer your first question; consider the operator \frac{d}{dx} which differentiates some expression to the right of it with respect to x, let's call this expression y.

    First derivative: \frac{d}{dx}(y)

    Seconds derivative: \frac{d}{dx}(\frac{d}{dx}(y))

    \displaystyle =\frac{d}{dx}(\frac{dy}{dx}) = \frac{d(\frac{dy}{dx})}{dx} = \frac{ddy}{dxdx} = \frac{d^2y}{dx^2}

    The reason why the d on the 'denominator' is not squared (I think, may get corrected on this) is because we are taking x twice for the second derivative but only considering one change. I'm not an expert on this notation so perhaps someone else can explain.
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    (Original post by RDKGames)
    To answer your first questions; consider the operator \frac{d}{dx} which differentiates some expression to the right of it with respect to x, let's call this expression y.

    First derivative: \frac{d}{dx}(y)

    Seconds derivative: \frac{d}{dx}(\frac{d}{dx}(y))

    =\frac{d}{dx}(\frac{dy}{dx}) = \frac{d(\frac{dy}{dx})}{dx} = \frac{ddy}{dxdx} = \frac{d^2y}{dx^2}

    The reason why the d on the 'denominator' is not squared (I think, may get corrected on this) is because we are taking x twice for the second derivative but only considering one change. I'm not an expert on this notation so perhaps someone else can explain.
    You're making the mistake of treating d as an actual variable. It's not. dx is an object, it's a thing in it's own right. It's just abbreviated using two letters instead of 1. So when you have dx dx = dx^2 that's the whole of dx being squared because the d and the x aren't two seperate things, they are one object dx. So dx^2 = dx * dx. It's sort of like you saying that sin^2 x means s * i * n * n x just becase the sqaure is on the n, it's on the entire object sin^2, since sin is an object in its own right, just like dx.

    Of course, the above is just intuition for why the notation should be like that, in reality, it's just notation and we can write it whatever way we want, there isn't any proper maths behind it.
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    (Original post by Zacken)
    You're making the mistake of treating d as an actual variable. It's not. dx is an object, it's a thing in it's own right. It's just abbreviated using two letters instead of 1. So when you have dx dx = dx^2 that's the whole of dx being squared because the d and the x aren't two seperate things, they are one object dx. So dx^2 = dx * dx. It's sort of like you saying that sin^2 x means s * i * n * n x just becase the sqaure is on the n, it's on the entire object sin^2, since sin is an object in its own right, just like dx.

    Of course, the above is just intuition for why the notation should be like that, in reality, it's just notation and we can write it whatever way we want, there isn't any proper maths behind it.
    Thanks for that example, couldn't find a way to think about it differently. I blame Leibnitz for this damn notation.
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    Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

    when f''(x) = 0 you need to investigate further right?
    so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

    my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

    i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.
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    (Original post by ihatePE)
    Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

    when f''(x) = 0 you need to investigate further right?
    so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

    my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

    i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.
    wot?

    for a stationary point to be an inflection point, you need f'(x) = 0 and f''(x) = 0.

    [and the obvious caveat for anybody who cares]
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    (Original post by ihatePE)
    Hello, i just want a confirmation relating to ''determining whether a stationary point is a point of inflection''

    when f''(x) = 0 you need to investigate further right?
    so what i do is the 'gradient table' where i find out what dy/dx is when x=-1 x=0 x=1 ect

    my point is, for it to be a point of inflection, dy/dx must be positive/negative and NOT change its sign (-/+) when x=-1 and x=1... am i right?

    i've read the notes in the book for ages and its worded so briefly that im not so sure if that's the rules.
    I'll have some of whatever drugs you're having, mate.

    Stationary points where f'(x)=0
    Points of inflection where f''(x)=0

    Combine the two for both to be true.

    "dy/dx must be positive/negative and NOT change its sign (-/+)" ...huh?
    "so what i do is the 'gradient table'..." ...why? Just solve for values of x then solve for values of x on f''(x)=0. Compare the two for the same x's and you've found the x coordinates of the points which are stationary AND points of inflection at the same time.
 
 
 
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