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    im stuck on two questions, any help appreciated

    1: given that y=8x^3 - 4x^0.5 + (3x^2 + 2)/x , x>0

    i know how to differentiate but what does the x>0 mean, is it there to throw me off?

    2: the curve c has equation y=(x+3)(x-8)/x , x>0
    a: find dy/dx in its simplest form
    b: find an equation of the tangent to c at the point where x=2

    i started working it out but then got a ridiculous answer

    thanks
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    (Original post by not_lucas)
    im stuck on two questions, any help appreciated

    1: given that y=8x^3 - 4x^0.5 + (3x^2 + 2)/x , x>0

    i know how to differentiate but what does the x>0 mean, is it there to throw me off?

    2: the curve c has equation y=(x+3)(x-8)/x , x>0
    a: find dy/dx in its simplest form
    b: find an equation of the tangent to c at the point where x=2

    i started working it out but then got a ridiculous answer

    thanks
    Please post your working for part 2

    if x = 0 then dividing by x will be a problem. It shouldn't matter for your working out, though.
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    i got dy/dx to = x-24x^-1 but i dont think it is right
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    (Original post by not_lucas)
    i got dy/dx to = x-24x^-1 but i dont think it is right
    Unfortunately not. I fear that you may have forgotten to apply the quotient rule properly, did you remember to
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    divide by the square of the denominator instead of just the denominator? Because your answer is nearly correct.
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    (Original post by not_lucas)
    i got dy/dx to = x-24x^-1 but i dont think it is right
    It's not correct. Can you post your working?
    You don't need to use the quotient rule for this. Just simplify the fractions.
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    i multiplied out the brackets to get x^2-5x-24 which i then divided by x as shown in the curve c equation then from there i ended up with x-5-24x^-1 which i differentiated to x-24x^-1. I think i know where i went partly wrong, the x should become 1 due to it being 1x shouldn't it?

    Also, im stuck on another question which asks me to find the area of a triangle, i have two coordinates but would i not need a third, i got those coordinates by subbing y=0 into the equations given as its says the normal and tangent meet the x axis at B and A respectively, do i get the third lot of coordinates by using simultaneous equations with the normal and tangent equations or something?
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    (Original post by not_lucas1)
    i multiplied out the brackets to get x^2-5x-24 which i then divided by x as shown in the curve c equation then from there i ended up with x-5-24x^-1 which i differentiated to x-24x^-1. I think i know where i went partly wrong, the x should become 1 due to it being 1x shouldn't it?
    Use the rule that \frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}, and \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)), so using this on your function we have

    \frac{\mathrm{d}}{\mathrm{d}x} (x - 24x^{-1}) = \frac{\mathrm{d}}{\mathrm{d}x} (x^1) - \frac{\mathrm{d}}{\mathrm{d}x} (24x^{-1}) = \cdots You should get
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    1 - 24x^{-2}
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    (Original post by not_lucas1)
    Also, im stuck on another question which asks me to find the area of a triangle, i have two coordinates but would i not need a third, i got those coordinates by subbing y=0 into the equations given as its says the normal and tangent meet the x axis at B and A respectively, do i get the third lot of coordinates by using simultaneous equations with the normal and tangent equations or something?
    You would be better off posting a new thread entirely with the full question. :-)
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    I got 1+ 24x^-2. If you times the -24 by the power of -1 shouldn't the result be positive?
 
 
 
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