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Math Help - Differentiation questions.

  1. #1
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    I am also on the next question, i have solved a few and struggling with others?

    2. obtain dy/dx for the following functions using the 'direct rule'

    a) y=2.5x^2
    =5x

    b) y=x^2/2.5
    =0.8x

    c) y=2.5x - 2.5
    =-2.5x^0

    d)y= (2.5)(SQUAREROOT OF x)
    =1.25x

    e)y=x^2.5
    =2.5^1.5

    f)y=2.5/x^2 HELP

    g)y=(^2.5 SQUARE ROOT OF x) HELP

    h)y=2.5x^4 - (x^3/2.5) - (x/2.5) + 2.5 SO FAR
    =10x^3 - 1.2x^2 - 0.4x^0 + 2.5

    I)y=2.5x{(x^2)-(1)} HELP

    J) y=(x^3 - 2.5x^2) / (2.5) (SQUARE ROOT OF x) HELP
    Last edited by mr fantastic; October 3rd 2010 at 12:23 PM. Reason: Moved to new thread.
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  2. #2
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    d)
    y = 2.5 \sqrt(x) = 2.5 x^{0.5}
    By the power rule
     y' = 0.5*2.5 x^{0.5-1} = \dfrac{1.25}{\sqrt{x}}

    f)
    y = \dfrac{2.5}{x^2} = 2.5 x^{-2} now use the power rule. [Typo Fixed]

    g) What?

    h) If C is a constant then \frac{d}{dx}[C] = 0
    I)
    y=2.5 x(x^2 -1) = 2.5x^3 - 2.5x Now use a combination of power rule and sum rule

    J)
    y=\dfrac{x^3 - 2.5x^2}{2.5\sqrt{ x}} = 0.4x^{2.5} - x^{1.5} Now use a combination of the power rule and sum rule.
    Last edited by lvleph; October 6th 2010 at 06:46 AM. Reason: Fixed Typo
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  3. #3
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    Ok i have attempted the questions you have helped me with:

    d) 1.25^-0.5x

    f) 5x

    h) 10x^3 - 1.2x - 0.4x + 2.5

    I) (7.5x^2) - (2.5x)

    J) (x^1.5) - (1.5x^0.5)

    Thankyou
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  4. #4
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    d) How did you get 1.25^-0.5x? lvleph had already given you the correct answer...


    e) y=x^{2.5}
    \frac{dy}{dx}=2.5^{1.5}

    Where'd your x go? If it's \frac{dy}{dx}=2.5x^{1.5} then it is correct.


    f) y=2.5/x^2
    Another way of writing this is y=2.5x^{-2}


    g) Doesn't make sense. Is it y= \sqrt{x}^{2.5})??


    h) y=2.5x^4 - \frac{x^3}{2.5} - \frac{x}{2.5} + 2.5
    You are almost there. Why do you still have that + 2.5 part at the end?


    i) (7.5x^2) - (2.5x)
    Almost there with this one as well. You haven't seemed to have differentiated the 2.5x part yet.


    j) Correct.
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  5. #5
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    Quote Originally Posted by lvleph View Post
    d)
    y = 2.5 \sqrt(x) = 2.5 x^{0.5}
    By the power rule
     y' = 0.5*2.5 x^{0.5-1} = \dfrac{1.25}{\sqrt{x}}

    f)
    y = \dfrac{2.5}{x^2} = 2.5 x^2 now use the power rule.
    Typo! lvleph surely meant 2.5 x^{-2}

    g) What?

    h) If C is a constant then \frac{d}{dx}[C] = 0
    I)
    y=2.5 x(x^2 -1) = 2.5x^3 - 2.5x Now use a combination of power rule and sum rule

    J)
    y=\dfrac{x^3 - 2.5x^2}{2.5\sqrt{ x}} = 0.4x^{2.5} - x^{1.5} Now use a combination of the power rule and sum rule.
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  6. #6
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    Maybe g) is  x^{1/2.5}?
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