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    (Original post by _Xenon_)
    Ah so was the answer: 0x^-1
    Thanks very much...
    Well, the answer is 0. Because 0 * anything = 0.
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    (Original post by Zacken)
    Well, the answer is 0. Because 0 * anything = 0.
    Ohh thanks. Will they ask these type of questions on IGCSE too?
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    (Original post by _Xenon_)
    Ohh thanks. Will they ask these type of questions on IGCSE too?
    Depends what IGCSE.

    Anyways, can you now do the original question in your first post? Please do it step-by-step like I did in my previous post! What do you get?
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    (Original post by Zacken)
    Depends what IGCSE.

    Anyways, can you now do the original question in your first post? Please do it step-by-step like I did in my previous post! What do you get?
    So dy/dx = 3*2x^3-1

    dy/dx = 6x^2 ?
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    (Original post by _Xenon_)
    So dy/dx = 3*2x^3-1

    dy/dx = 6x^2 ?
    We want to differentiate y = 2x^3 + x^2 - 8x + 3.

    So, you've correctly differentiated the first term, well done. Now what about the rest?
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    (Original post by Zacken)
    We want to differentiate y = 2x^3 + x^2 - 8x + 3.

    So, you've correctly differentiated the first term, well done. Now what about the rest?
    I don't know the rest this is hard
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    (Original post by _Xenon_)
    I don't know the rest this is hard
    Why aren't you doing it step-by-step like I told you do?

    If we want to differentiate the sum of all those things, we can just differentiate each term separately and then add them up.

    SO, to differentiate y = 2x^3 + x^2 - 8x + 3, you need to be able to differentiate each bit seperately:

    (i) differentiate 2x^3, you've done this correctly.
    (ii) differentiate x^2, you can do this, c'mon!
    (iii) differentiate -8x, you can do this!
    (iv) differentiate 3, what did we say the derivative of a constant is?
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    (Original post by Zacken)
    Why aren't you doing it step-by-step like I told you do?

    If we want to differentiate the sum of all those things, we can just differentiate each term separately and then add them up.

    SO, to differentiate y = 2x^3 + x^2 - 8x + 3, you need to be able to differentiate each bit seperately:

    (i) differentiate 2x^3, you've done this correctly.
    (ii) differentiate x^2, you can do this, c'mon!
    (iii) differentiate -8x, you can do this!
    (iv) differentiate 3, what did we say the derivative of a constant is?
    Oh I think I can do it and then do I put a + sign in between them all
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    (Original post by _Xenon_)
    Oh I think I can do it and then do I put a + sign in between them all
    Yep - pretty much. What do you get?

    By the way, you might want to watch this as a very detailed example. Please take some time to sit down and understand all of this.
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    (Original post by Zacken)
    Yep - pretty much. What do you get?

    By the way, you might want to watch this as a very detailed example. Please take some time to sit down and understand all of this.
    Thanks very much - I'm very tired atm so cannot concentrate very well atm.
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    (Original post by _Xenon_)
    Thanks very much - I'm very tired atm so cannot concentrate very well atm.
    Come back and read this thread again in the morning.
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    (Original post by Zacken)
    Come back and read this thread again in the morning.
    Yep, thanks. Maybe I should sleep early and wake up early I need to revise science too so much work lol. Thanks for your help .
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    (Original post by _Xenon_)
    Ah so was the answer: 0x^-1
    Thanks very much...
    It may help you to bear in mind that dy/dx gives you a formula for the slope/gradient/steepness of a function at any point. If you start with, say:

    y=3x

    then you already know the gradient, since that's just the equation of a straight line through the origin i.e. it's in the form y=mx+c with m being the gradient. So you can see that:

    y=3x \Rightarrow \frac{dy}{dx} =3

    without relying on any rules. Also, since y=c is the equation of a straight line, parallel to the x-axis, at height c, then you know that it has no rise or fall anywhere, so its gradient is 0 everywhere. So you can also see that:

    y=c \Rightarrow \frac{dy}{dx} = 0

    without relying on any rules. Once that is clear in your mind, you can follow the suggestions above to see how the rules give the right answers in these two cases.
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    (Original post by Zacken)
    Okay, here are some rules that you should get used to for differentiating. What \frac{dy}{dx} is is a linear operator. It's read as "the derivative of y with respect to x". Because it's linear, it's a very easy and nice operator.

    So, we have, intuitively:

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }}{\mathrm{d}x}(f(x) + g(x)) = \frac{\mathrm{d}}{\mathrm{d}x}(f  (x)) + \frac{\mathrm{d}}{\mathrm{d}x} (g(x))end{equation*}

    All this is saying that if you want to differentiate the sum of two (or more!) terms, you can differentiate each term separately and then add them together.

    So, for example - you know that when you differentiate x^1, you get 1x^0 = 1. So if you were to differentiate 2x you can simply write 2x = x+x and then it follows that \frac{d}{dx}(2x) = \frac{\mathrm{d}}{\mathrm{d}x}(x  ) + \frac{\mathrm{d}}{\mathrm{d}x}(x  ) = 1 + 1 = 2. Which makes sense!

    Following on from that rule, it follows just as intuitively that

    \displaystyle 

\begin{equation*}\frac{\mathrm{d  }}{\mathrm{d}x}(\lambda x) = \lambda \frac{\mathrm{d}}{\mathrm{d}x}(x  )\end{equation*}.

    That is, if you want to differentiate a constant multiple of x, you can pull the multiple out, differentiate the x term and then put the constant back in. This makes total sense for differentiating 2x because we can do \frac{\mathrm{d}}{\mathrm{d}x}(2  x) = 2\frac{\mathrm{d}}{\mathrm{d}x}(  x) = 2 (1) = 2 just like last time!

    Now, you've just learnt that if you want to differentiate x raised to a power, you do this:

    \displaystyle

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x}(x  ^n) = nx^{n-1} \end{equation*}

    i.e: all you need to do is multiply the entire thing by the power and then subtract 1 from the power. Do you understand this? Please tell me if you don't!

    With these three rules, you can differentiate pretty much anything! :eek3: How cool is that?! :woo:

    If you want to differentiate a polynomial, you just differentiate each term separately and then add them up together.

    So - for example, if you want to differentiate y = 2x + 4x^2, I'd advise you to do this one step at a time (for now, once you get the hang of it, you can skip steps).

    But for now, I want you to do something like this:

    \displaystyle

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x}(2  x + 4x^2) \\  &= \frac{\mathrm{d}}{\mathrm{d}x}(2  x) + \frac{\mathrm{d}}{\mathrm{d}x}(4  x^2) \\ & = 2\frac{\mathrm{d}}{\mathrm{d}x}(  x) + 4\frac{\mathrm{d}}{\mathrm{d}x}(  x^2) \\ & = 2\frac{\mathrm{d}}{\mathrm{d}x}(  x^1) + 4\frac{\mathrm{d}}{\mathrm{d}x}(  x^2) \\ & = 2(1 \times x^{1-1}) + 4(2 \times x^{2-1}) \\ &= 2(1 \times x^0) + 4(2 \times x) \\ &= 2(1) + 4(2x) \\ & = 2 + 8x \end{align*}

    Please point out anything that you didn't understand in here.

    Can you do these for me:

    y = 3x + 4
    y = 3x^2 + 5x
    OK is the first one 3?
    Is the second 6x+5?
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    (Original post by _Xenon_)
    OK is the first one 3?
    Is the second 6x+5?
    Both are correct.

    Some harder ones if you want to try them :

    y=2x^2 - 3

    y = 3x^2 - 2x + 7

    y = 5x^4 - 2x^2 + 3
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    (Original post by _Xenon_)
    OK is the first one 3?
    Is the second 6x+5?
    Yes and yes.
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    (Original post by notnek)
    Both are correct.

    Some harder ones if you want to try them :

    y=2x^2 - 3

    y = 3x^2 - 2x + 7

    y = 5x^4 - 2x^2 + 3
    1.) 4x
    2.) 6x-2
    3.) 20x^3-4x



    (Original post by Zacken)
    Yes and yes.
    Thanks I think I understand this now!
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    (Original post by _Xenon_)
    1.) 4x
    2.) 6x-2
    3.) 20x^3-4x
    Very well done!

    Notnek deserves most of the thanks.
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    (Original post by _Xenon_)
    1.) 4x
    2.) 6x-2
    3.) 20x^3-4x





    Thanks I think I understand this now!
    Can you tell me what qualification you are doing? E.g. IGCSE Edexcel?

    Then we will know if you will need to practice harder questions or not.
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    (Original post by notnek)
    Can you tell me what qualification you are doing? E.g. IGCSE Edexcel?

    Then we will know if you will need to practice harder questions or not.
    Hi yes thanks.
    I'm doing IGCSE Edexcel.
 
 
 
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