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Differentiate from first principles? watch

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    Hey guys, I have been set 5 questions for homework, and I know how to different and all, but I don't know how to differentiate "from first principles", so can anybody help to use this method to answer these questions for me step by step? Would be really appreciated

    1) y = x^2 + 2
    2) y = x^2 - x
    3) y = 4x
    4) y = x - 1
    5) y = x^3

    Many thanks
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    For all those just work out

    [y(x+h) - y(x)]/h

    simplify, and then set h to be zero
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    Differentiation from first principles means that for each of the functions, you calculate \dfrac{f(x+h) - f(x)}{h} (where y=f(x)) and then observe what happens as h \to 0.

    I'll get you started on the first of your problems, and then you can do the rest. You have y=x^2+2, and so we need to look at
    \dfrac{((x+h)^2 + 2) - (x^2+2)}{h}

    Expanding the (x+h)² bracket on the numerator gives
    \dfrac{x^2+2hx+h^2+2-x^2-2}{h}

    ...and simplifying reveals that this is equal to
    2x+h

    Clearly as h tends to zero, this tends to 2x.

    Now you do the rest!
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    (Original post by adil12)
    Hey guys, I have been set 5 questions for homework, and I know how to different and all, but I don't know how to differentiate "from first principles", so can anybody help to use this method to answer these questions for me step by step? Would be really appreciated

    1) y = x^2 + 2
    2) y = x^2 - x
    3) y = 4x
    4) y = x - 1
    5) y = x^3

    Many thanks
    If you're differentiating a function f(x) from first principle you ned to express it's derivative as follows:
    f'(x) = \displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
    and then manipulate it until you get the correct answer.
    So for the first one, f(x)=x^2+2, therefore f(x+h) = (x+h)^2+2. Put them into that above expression and manipulate it until you don't have the fraction with a denominator which tends to zero.
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    (Original post by adil12)
    Hey guys, I have been set 5 questions for homework, and I know how to different and all, but I don't know how to differentiate "from first principles", so can anybody help to use this method to answer these questions for me step by step? Would be really appreciated

    1) y = x^2 + 2
    2) y = x^2 - x
    3) y = 4x
    4) y = x - 1
    5) y = x^3

    Many thanks
    You have to use the definition of differentiation:

    f'(x) = \lim_{h \to o} \frac{f(x+h)-f(x)}{h}

    Edit: 4 good replies in a minute. Speedy stuff guys :P
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    (Original post by nuodai)
    Differentiation from first principles means that for each of the functions, you calculate \dfrac{f(x+h) - f(x)}{h} (where y=f(x)) and then observe what happens as h \to 0.

    I'll get you started on the first of your problems, and then you can do the rest. You have y=x^2+2, and so we need to look at
    \dfrac{((x+h)^2 + 2) - (x^2+2)}{h}

    Expanding the (x+h)² bracket on the numerator gives
    \dfrac{x^2+2hx+h^2+2-x^2-2}{h}

    ...and simplifying reveals that this is equal to
    2x+h

    Clearly as h tends to zero, this tends to 2x.

    Now you do the rest!
    I see what you did here, clear and productive
    Thanks a lot! Can you tell me how I can give you rep?
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    (Original post by PhyMath)
    You have to use the definition of differentiation:

    f'(x) = \lim_{h \to o} \frac{f(x+h)-f(x)}{h}

    Edit: 4 good replies in a minute. Speedy stuff guys :P
    Wow you guys replied so fast all with the same good answers, thanks to all for help And thanks mate, got it now
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    (Original post by Farhan.Hanif93)
    If you're differentiating a function f(x) from first principle you ned to express it's derivative as follows:
    f'(x) = \displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
    and then manipulate it until you get the correct answer.
    So for the first one, f(x)=x^2+2, therefore f(x+h) = (x+h)^2+2. Put them into that above expression and manipulate it until you don't have the fraction with a denominator which tends to zero.
    Thanks mate, solid stuff
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    (Original post by RichE)
    For all those just work out

    [y(x+h) - y(x)]/h

    simplify, and then set h to be zero
    Thanks for the help man! Helped out fast and well :p:
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    (Original post by RichE)
    For all those just work out

    [y(x+h) - y(x)]/h

    simplify, and then set h to be zero
    Set h to be zero?!

    That's ******* appalling, that completely detracts from the whole idea behind calculus.

    Newton/Liebniz would be turning in his grave.
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    (Original post by RichE)
    For all those just work out

    [y(x+h) - y(x)]/h

    simplify, and then set h to be zero
    Set h to be zero?!

    That's ******* appalling, that completely detracts from the whole idea behind calculus.

    Newton/Liebniz would be turning in his grave.
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    (Original post by adil12)
    I see what you did here, clear and productive
    Thanks a lot! Can you tell me how I can give you rep?
    Pleasure Rep can be given by clicking the image at the top-right of a post which has the thumbs up/down in it.
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    (Original post by nuodai)
    Pleasure Rep can be given by clicking the image at the top-right of a post which has the thumbs up/down in it.
    Repped! By the way can you help me with the second question? I can do the rest but I don't know how to go about this one
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    (Original post by adil12)
    Repped! By the way can you help me with the second question? I can do the rest but I don't know how to go about this one
    If you can show some working, we'll be able to give you a better hint.
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    (Original post by turnand)
    Set h to be zero?!

    That's ******* appalling, that completely detracts from the whole idea behind calculus.

    Newton/Liebniz would be turning in his grave.
    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
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    (Original post by Farhan.Hanif93)
    If you can show some working, we'll be able to give you a better hint.
    Alright mate, I used the general formula and then got:
    ((x+h)^2 - x) - (x^2 - x)/h
    = x^2 + 2hx + h^2 - x - x^2 + x /h
    = 2x + h, and since h tends to be 0, answer is 2x, which I know is wrong because if I differentiate with powers I know it should be 2x-1 as the answer, where have I gone wrong?
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    (Original post by adil12)
    Alright mate, I used the general formula and then got:
    ((x+h)^2 - x) - (x^2 - x)/h
    = x^2 + 2hx + h^2 - x - x^2 + x /h
    = 2x + h, and since h tends to be 0, answer is 2x, which I know is wrong because if I differentiate with powers I know it should be 2x-1 as the answer, where have I gone wrong?
    If f(x) = x^2-x then f(x+h) \ne (x+h)^2 -x
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    (Original post by adil12)
    Alright mate, I used the general formula and then got:
    ((x+h)^2 - x) - (x^2 - x)/h
    = x^2 + 2hx + h^2 - x - x^2 + x /h
    = 2x + h, and since h tends to be 0, answer is 2x, which I know is wrong because if I differentiate with powers I know it should be 2x-1 as the answer, where have I gone wrong?
    f(x+h) = (x+h)^2-(x+h) in this case...
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    (Original post by nuodai)
    If f(x) = x^2-x then f(x+h) \ne (x+h)^2 -x
    But that's what I done no?
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    (Original post by adil12)
    But that's what I done no?
    The \ne means "not equal to", but you subbed it in as if it were equal to it. It should have been as in Farhan's post (#18).
 
 
 
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