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    (Original post by Farhan.Hanif93)
    f(x+h) = (x+h)^2-(x+h) in this case...
    Then it expands to:
    x^2 + 2hx + h^2 - x - h - x^2 + x /h?
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    (Original post by adil12)
    Then it expands to:
    x^2 + 2hx + h^2 + x + h - x^2 + x / h
    How can you simplify when the you have something like 2x/h as a term?
    Not quite, you've got a sign error where I've highlighted bold. Those two terms should be subtracted.
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    (Original post by Farhan.Hanif93)
    Not quite, you've got a sign error where I've highlighted bold. Those two terms should be subtracted.
    My bad, corrected it now and simplifies to:
    2x + h -1 and since h tends to be 0, it's 2x -1
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    (Original post by adil12)
    My bad, corrected it now and simplifies to:
    2x + h -1 and since h tends to be 0, it's 2x -1
    :yes:
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    (Original post by Farhan.Hanif93)
    :yes:
    Thanks a lot mate! Helped a lot
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    (Original post by RichE)
    I said it would work for those - which it does - and which is pretty much how Newton did it originally - which rather makes your post (C) Bishop Berkeley
    Pretty much how newton did it originally?

    I have no respect that at all, read the paper before you make comments about stuff you obviously haven't seen.
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    (Original post by turnand)
    Pretty much how newton did it originally?

    I have no respect that at all, read the paper before you make comments about stuff you obviously haven't seen.
    Even though you're just starting uni now and he's a graduate... :rolleyes:
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    (Original post by Farhan.Hanif93)
    Even though you're just starting uni now and he's a graduate... :rolleyes:
    It doesn't matter, he could be steven hawking, he's still wrong
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    (Original post by turnand)
    Pretty much how newton did it originally?

    I have no respect that at all, read the paper before you make comments about stuff you obviously haven't seen.
    Old proverb: when in a hole - stop digging.
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    (Original post by turnand)
    It doesn't matter, he could be steven hawking, he's still wrong
    If you still feel that way in a month, then don't bother coming along to my Analysis I lectures.

    More specifically to the questions at hand:

    (i) There's nothing wrong with what I wrote in my original post, and I don't simply mean that they produce the right answer, there's nothing wrong mathematicallly with what I said. Nothing I wrote warranted expletives.

    If you take the y = x^2 + 2 example then:

    [y(x+h)-y(x)]/h = [(x+h)^+2 - x^2 - 2]/h = (2xh+h^2)/h = 2x + h

    and putting h = 0 gives 2x. This is the correct answer, but more than that: in any situation where one has simplified [y(x+h)-y(x)]/h as a function f(h) which is defined and continuous at 0, then f(0) is the correct answer as  f(0) = \lim_{h->0} f(h).

    (ii) Pedagogically, what was so wrong with solving the question this way for someone who is just meeting differentiation? Do you think it would have been clearer to the OP if I'd said "you then let h tend to 0"? Do you know what that means yourself, in a well-quantified manner? Even if you do, sharing that with the OP would have just baffled them. Even saying "you let h become small" (which is how this is often paraphrased) is vague and why would the answer not be 2x + small? Why not let the OP get these questions under their belt and then worry about that aspect when they meet y = sinx or y = e^x?

    (iii) And finally, this is pretty much what Newton did. Newton's calculus wasn't rigorous and was (posthumously) rightly criticised by Bishop Berkeley for what Berkeley called the "ghosts of departed quantities". Newton's method of fluxions involved an "o" notation which would disappear at the end of the calculation once it didn't cause any issues to do so. Rigour was still an issue in the analysis of Cauchy and Fourier. It wasn't until Weierstrass in the nineteenth century that there was a rigorous definition of limit.
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    (Original post by turnand)
    I have no respect that at all, read the paper before you make comments about stuff you obviously haven't seen.
    :laugh: Pretty sure RichE would have seen this before. Possibly read Newton's papers as well.


    It might be nice to point out to the OP that setting h=0 is legitimate because you have removed the problem of dividing by 0 when simplifying.
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    (Original post by RichE)
    ...stuff...
    You had me at "my Analysis I lectures".
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    Bump.

    I've got no problem with some of the 'harder' examples of this I've been given, but I wasn't sure how to approach f(x) = 7.

    Normally you replace the x in the function with (x + δx), but what would I do here?
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    (Original post by graves248)
    Bump.

    I've got no problem with some of the 'harder' examples of this I've been given, but I wasn't sure how to approach f(x) = 7.

    Normally you replace the x in the function with (x + δx), but what would I do here?
    If f(x) = 7 for all x then it's also 7 if you replace x with (x + δx).
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    Wouldn't that give you a newton quotient of:

    (f(x + δx) - f(x))/δx

    ie 0/δx ?

    Since I can't use l'hopitals rule to find that limit, I'm a bit lost as to how I get an answer of 0.
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    (Original post by graves248)
    Wouldn't that give you a newton quotient of:

    (f(x + δx) - f(x))/δx

    ie 0/δx ?

    Since I can't use l'hopitals rule to find that limit, I'm a bit lost as to how I get an answer of 7.
    You don't need L'Hopital. Remember, you haven't taken the limit yet so\displaystyle\lim_{\delta x \to 0} 0/\delta x = \displaystyle\lim_{\delta x \to 0} 0

    Also, why are you looking for an answer of 7? I think you'll find that \dfrac{d}{dx}7 \not=7. :p:
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    Yeah, not a clue why I typed that. Cheers for the help, I just thought that you couldn't take the limit of something divided by zero.
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    (Original post by graves248)
    Yeah, not a clue why I typed that. Cheers for the help, I just thought that you couldn't take the limit of something divided by zero.
    This is the important thing to realise, \delta x is not zero. it's simply a quantity that we intend to make small eventually (in this case).

    Intuitively, you could take the limit without using what I've said above directly by realising that nothing can approach 0 'faster' than 0 itself so the numerator of the expression heads to zero 'more rapidly' than δx and hence the whole expression approaches zero in the limit.
 
 
 
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