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Math Help - Differentiation from 1st principles

  1. #1
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    Differentiation from 1st principles

    I don't know how to do this but I'd like to learn how:

    A question is "Given that 2x - 5x + 3, find dy/dx from first principles."

    Can anyone please help me
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  2. #2
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    Quote Originally Posted by db5vry View Post
    I don't know how to do this but I'd like to learn how:

    A question is "Given that 2x - 5x + 3, find dy/dx from first principles."

    Can anyone please help me
    Hmmmm....

    dy/dx = 4x - 5 ?

    To learn how to differentiate polynomials.
    http://en.wikipedia.org/wiki/Calculus_with_polynomials
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  3. #3
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    Quote Originally Posted by TitaniumX View Post
    Hmmmm....

    dy/dx = 4x - 5 ?

    To learn how to differentiate polynomials.
    Calculus with polynomials - Wikipedia, the free encyclopedia
    Yeah I know how to find dy/dx, you just multiply 2 by the and get 4 and reduce the power, then minus the 5.
    But does anyone know how to do this via first principles?
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  4. #4
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    it's just a fancy way of saying plugging it into the definition of a derivative in terms of the limit as x approaches 0.
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  5. #5
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    Quote Originally Posted by db5vry View Post
    I don't know how to do this but I'd like to learn how:

    A question is "Given that 2x - 5x + 3, find dy/dx from first principles."

    Can anyone please help me
    Write out the difference quotient for f(x)=2x^2-5x+3:

     <br />
DQ(f,h)=\frac{f(x+h)-f(x)}{h}=\frac{2(x+h)^2-5(x+h)+3-2x^2+5x-3}{h}<br />

    .............. = \frac{4hx+2h^2-5h}{h}=4x-5+2h

    So:

     <br />
\frac{df}{dx}=\lim_{h \to 0} DQ(f,h)=4x-5<br />

    CB
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