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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.
    how can I differentiate with respect to x

    1/x
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    (Original post by _Xenon_)
    how can I differentiate with respect to x

    1/x
    Rewrite it as \frac{1}{x} = x^{-1} and then it's the same thing all over again.
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    (Original post by Zacken)
    Rewrite it as \frac{1}{x} = x^{-1} and then it's the same thing all over again.
    -x^-2 ?
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    (Original post by _Xenon_)
    -x^-2 ?
    Indeed. And can you write that in fraction form for the bantz?
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    (Original post by Zacken)
    Indeed. And can you write that in fraction form for the bantz?
    not sure how ?
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    (Original post by _Xenon_)
    not sure how ?
    x^-n = 1/(x^n)
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    I've linked a set of IGCSE Edexcel calculus questions to give an idea of how hard the questions will be.

    http://www.speedyshare.com/g9sEV/dydx.pdf

    I won't be around much this afternoon so hopefully Zacken can help you with some of these if he's free.

    Zacken, I think you did CIE IGCSE? So I don't know if you are aware of the IGCSE spec but these questions should help.
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    (Original post by notnek)
    I've linked a set of IGCSE Edexcel calculus questions to give an idea of how hard the questions will be.

    http://www.speedyshare.com/g9sEV/dydx.pdf

    I won't be around much this afternoon so hopefully Zacken can help you with some of these if he's free.

    Zacken, I think you did CIE IGCSE? So I don't know if you are aware of the IGCSE spec but these questions should help.
    Cheers, thanks for the link, just downloaded it. I'll be around for most of the evening, so I should be able to help. _Xenon_ - mind downloading the pdf he's linked and then we can go through the questions together? Post a picture of your answers or ask for help with any of them (I've got a copy, so you just need to reference the question number) and I can verify your answers and/or help out with troubles you have.
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    (Original post by Zacken)
    Cheers, thanks for the link, just downloaded it. I'll be around for most of the evening, so I should be able to help. _Xenon_ - mind downloading the pdf he's linked and then we can go through the questions together? Post a picture of your answers or ask for help with any of them (I've got a copy, so you just need to reference the question number) and I can verify your answers and/or help out with troubles you have.
    OK thanks man.
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    (Original post by _Xenon_)
    OK thanks man.
    No worries, when do you want to start?
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    (Original post by Zacken)
    No worries, when do you want to start?
    just downloaded and now I'll try them, thanks.
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    (Original post by _Xenon_)
    just downloaded and now I'll try them, thanks.
    Awesome - just post your answers here and I'll verify them whenever I can!
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    (Original post by Zacken)
    Awesome - just post your answers here and I'll verify them whenever I can!
    First page:

    http://i.imgur.com/b7gH0HX.jpg
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    Firstly, notation wise, on the last question y is not =3X^2 - Y=x^3. You want to write is as dy/dx =3x^2, or you can also write y'=3x^2. Once you have the expression for the gradient, you should be able to work out the values of x that satisfy the conditions, and then you can use the original equation (y=x^3 to get the coordinates)
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    First two is correct, well done!

    The gradient at any point (a,b) on a curve y = f(x) is given by finding \frac{\mathrm{d}y}{\mathrm{d}x} \bigg|_{x=a}.

    So, for example, the gradient at (3,9) on the curve y = x^2 is given by me differentiating the equation of the curve to get 2x and then plugging x=3 into this derivative to get gradient at (3,9) = 2 \times 3 = 6.

    Now, if I want to find the point at which the gradient of the curve is a certain value, I just take my derivative and set it equal to that value and solve the equation. So, for example, if I want the point at which the gradient of y=x^2 is 12 then I take the derivative 2x (this is my gradient at any point x), but I want the gradient to be 12, so 2x = 12 \Rightarrow x = 6.

    This is the x-coordinate of the point at which the gradient is 12; the y-coordinate is given by plugging my x into my function to get y = 6^2 = 36. So the point (6, 36) on the curve y = x^2 has gradient 12.

    What is the gradient at (2, 8) of the curve y = x^3?

    What point of the curve y=x^3 has gradient 27?

    Once you've done this for me, can you now do the question in the booklet?

    But yeah, when you find the derivative, you should not be doing:

    y = x^3
    y = 3x^2

    You should do:

    y=x^3
    dy/dx = 3x^2

    They mean very different things and you should be careful as to how you write them.
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    (Original post by samb1234)
    Firstly, notation wise, on the last question y is not =3X^2 - Y=x^3. You want to write is as dy/dx =3x^2, or you can also write y'=3x^2. Once you have the expression for the gradient, you should be able to work out the values of x that satisfy the conditions, and then you can use the original equation (y=x^3 to get the coordinates)
    Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

    So 2^3 = 8? Then what
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    (Original post by _Xenon_)
    Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

    So 2^3 = 8? Then what
    So one point is (2, 8)

    Now, do you remember how to solve equations of the form x^2 = 4? One solution is x = \sqrt{4} = 2. What about the other solution?

    Edit: What two numbers can you square to give you 4?
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    (Original post by _Xenon_)
    Oops that was what I meant, sorry. I was supposed to write dy/dx but wrote y.

    So 2^3 = 8? Then what
    You're missing one value of x...
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    (Original post by Zacken)
    So one point is (2, 8)

    Now, do you remember how to solve equations of the form x^2 = 4? One solution is x = \sqrt{4} = 2. What about the other solution?

    Edit: What two numbers can you square to give you 4?
    Is the other one negative? So (-2,-8)
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    (Original post by _Xenon_)
    Is the other one negative? So (-2,-8)
    Yep! Well done.
 
 
 
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