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    The gradient of a curve is 6x^2 + 12x^1/2. The curve passes through the points (4,10). Find the equation of the curve. What would i do from start to finish ?


    Also another question when differentiation this x - 3/x^2, Would do i do with the 3/x^2?
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    (Original post by Timmy12321)
    The gradient of a curve is 6x^2 + 12x^1/2. The curve passes through the points (4,10). Find the equation of the curve. What would i do from start to finish ?
    If you know the gradient, you can integrate it to find the equation of the curve. When you integrate, you will get an arbitrary constant C, which you should work out using the fact that the curve passes through the point (4,10).

    Also another question when differentiation this x - 3/x^2, Would do i do with the 3/x^2?
    Writing \dfrac{3}{x^2} = 3x^{-2} should help.
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    (Original post by Timmy12321)
    Also another question when differentiation this x - 3/x^2, Would do i do with the 3/x^2?
    I used to have trouble with these kinds of problems. He's right, writing \dfrac{3}{x^2} = 3x^{-2} should help but in order to understand why, you should probably re-visit Indices from Core 1. The main thing you need to know though is that \dfrac{1}{x} = x^{-1} and so \dfrac{1}{x^2} = x^{-2} The only difference now is when you have a numerator but if you think about \dfrac{1}{x^2} = 1x^{-2} then it is easy to work out.
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    (Original post by Timmy12321)
    The gradient of a curve is 6x^2 + 12x^1/2. The curve passes through the points (4,10). Find the equation of the curve. What would i do from start to finish ?

    Also another question when differentiation this x - 3/x^2, Would do i do with the 3/x^2?
    This is an integration question. 6x^2 + 12x^(1/2) becomes 2x^3 + 8x^(3/2) + c = y. Remember, add one to the power, divide by the new power. To find the full equation, sub in your given points, (4,10) 2(4)^3+8(4)(2)+c=10 which becomes 128+64+c=10 c=-182

    Therefore the answer is 2x^3 + 8x^(3/2) - 182 = y

    The second question is the same as asking for dy/dx of x-3x^(-2). The reverse rule to above applies, nax^(n-1) is the differential. So dy/dx=1+6/(x^3) or 1+6x^(-3). Because -3x x -2 = 6x -2-1=-3

    Or at least I hope I'm right.


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    Thanks guys, one last question what about this differentiation 20 x^4 + 6x^-3/2
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    (Original post by Timmy12321)
    Thanks guys, one last question what about this differentiation 20 x^4 + 6x^-3/2
    I think 80x^3 - 9x^(-2.5), but don't take my word for it.


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    20 x^4 + 6x^-3/2

    dy/dx = 80x^3 - 9x^-5/2
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    (Original post by Timmy12321)
    The gradient of a curve is 6x^2 + 12x^1/2. The curve passes through the points (4,10). Find the equation of the curve. What would i do from start to finish ?


    Also another question when differentiation this x - 3/x^2, Would do i do with the 3/x^2?
    You're given f'(x). Integrate and you will get f(x), but remember there is a constant that needs finding. f(4)=10 and now find the constant.

    I swear this a C1 question.
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    (Original post by Angryification)
    You're given f'(x). Integrate and you will get f(x), but remember there is a constant that needs finding. f(4)=10 and now find the constant.

    I swear this a C1 question.
    It became a C2 question the minute you were given coordinates technically. Before that it is C1 indefinite integration.


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    (Original post by Cynical_Smile01)
    It became a C2 question the minute you were given coordinates technically. Before that it is C1 indefinite integration.


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    You can be given coordinates in C1.
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    (Original post by Angryification)
    You can be given coordinates in C1.
    Depends what exam board.


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    (Original post by Cynical_Smile01)
    Depends what exam board.


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    I think OCR don't do it in C1 and keep finite integration to C2
 
 
 
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