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    How would I go about differentiating this? I'm confused because of the fraction:

    3x^2 + 7/2x
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    7/2 X^-1 takes the fraction away I think? I don’t know if that helps you
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    (Original post by CML2000)
    7/2 X^-1 takes the fraction away I think? I don’t know if that helps you
    By that I mean the X can be taken up top, not that the fraction completely disappears sorry 😅
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    (Original post by thedecorator33)
    How would I go about differentiating this? I'm confused because of the fraction:

    3x^2 + 7/2x
    Hi there :wavey:
    I've moved your thread to the Maths forum where hopefully you'll get more answers In future, it's best to select the subject from the list here and create a thread in there instead :yep: If you want to see where threads belong, check out this thread

    The fraction doesn't change anything as it's just a coefficient. You'd feel confident differentiating 3x^2 + 7x
    It's exactly the same method for 3x^2 + \frac{7}{2}x

    Unless the you mean the fraction is 3x^2 + \frac{7}{2x}? in which case you need to put it in terms of x^{-1} then differentiate as normal
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    (Original post by thedecorator33)
    How would I go about differentiating this? I'm confused because of the fraction:

    3x^2 + 7/2x
    To start, you should start by writing it properly. Is it 3x^2 + \frac{7}{2}x or \frac{3x^2+7}{2x} or 3x^2 + \frac{7}{2x} or...?
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    (Original post by Zacken)
    To start, you should start by writing it properly. Is it 3x^2 + \frac{7}{2}x or \frac{3x^2+7}{2x} or 3x^2 + \frac{7}{2x} or...?

    it's
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    (Original post by thedecorator33)
    There was no other way of me writing it differently. Its '7 over 2x' so 2x is the denominator while 7 is the numerator.
    So... 3x^2 + [7/(2x)]

    Anyway you know that the derivative of a sum is the sum of derivatives and the derivative of x^n is nx^(n-1) and 7/(2x) = (7/2) * x^(-1) and the derivative of af(x) is a f'(x).
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    (Original post by Lemur14)
    Hi there :wavey:
    I've moved your thread to the Maths forum where hopefully you'll get more answers In future, it's best to select the subject from the list here and create a thread in there instead :yep: If you want to see where threads belong, check out this thread

    The fraction doesn't change anything as it's just a coefficient. You'd feel confident differentiating 3x^2 + 7x
    It's exactly the same method for 3x^2 + \frac{7}{2}x

    Unless the you mean the fraction is 3x^2 + \frac{7}{2x}? in which case you need to put it in terms of x^{-1} then differentiate as normal
    Thank you for the reponse. It is the second one. How would I go about putting it in term of x^{-1}?
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    (Original post by thedecorator33)
    Thank you for the reponse. It is the second one. How would I go about putting it in term of x^{-1}?
    So you know that x^{-1} = \frac{1}{x}
    So what would be the value of n if \frac{7}{2x} = n \times \frac{1}{x}?
    Then you can write it as nx^{-1}
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    (Original post by Lemur14)
    So you know that x^{-1} = \frac{1}{x}
    So what would be the value of n if \frac{7}{2x} = n \times \frac{1}{x}?
    Then you can write it as nx^{-1}
    Ah so n would be 7/2
    Thanks for the helpful response.
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    (Original post by thedecorator33)
    Ah so n would be \frac{7}{2}

    Ah so n would be 7/2
    Thanks for the helpful response
    Yep n would be \frac{7}{2}
    Then you should be able to differentiate as normal
    No problem, glad I could help
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    (Original post by Zacken)
    So... 3x^2 + [7/(2x)]

    Anyway you know that the derivative of a sum is the sum of derivatives and the derivative of x^n is nx^(n-1) and 7/(2x) = (7/2) * x^(-1) and the derivative of af(x) is a f'(x).
    I think this is a bit too complicated for my level haha. Thanks for the help anyways.
 
 
 
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