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Derivitives- Calculus

  1. Jul 17, 2006 #1
    Hi
    I need help with the Derivites please.


    Find the value of dy/dx for the given value of x.

    y= (3-2x-x^2) (x^2+x-2), x=-2

    I have tried that I cant seem to understand the real steps to get the answer.
    I know I have to take the deriviate of this actual function...
    once i do that..I am not sure what to do next? plz help!
     
  2. jcsd
  3. Jul 17, 2006 #2

    Office_Shredder

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    Gold Member

    Once you have the derivative, just plug in x.

    Or are you having trouble taking the derivative?
     
  4. Jul 17, 2006 #3

    HallsofIvy

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    y is a product of two functions, 3- 2x- x2 and x2+ x- 2 so use the product rule: (fg)'= f 'g+ fg'. Or just go ahead and multiply it to get a single polynomial. Do you know how to find the derivative of a polynomial?
     
  5. Jul 17, 2006 #4
    no i dont know how to take the derivative of a polynomial?

    is that? let me try though.
    first i expand this two terms:
    i ended with some cubes nd quartics
    then i took the derivative.(when u multiply the exponent with coffiencent and the subtract the exponent with 1)

    thats wehre i get stuck. i need help here..

    the middle step is difficult..
    last part is easy when we plug x value..
     
  6. Jul 24, 2006 #5
    [tex]
    y = (-x^2-2x+3)(x^2+x-2)
    [/tex]

    let:
    [tex]
    u = f(x) = (-x^2-2x+3)
    [/tex]
    [tex]
    v = g(x) = (x^2+x-2)
    [/tex]
    [tex]
    y = uv
    [/tex]

    [tex]
    y + dy = (u + du)(v + dv) = uv + udv + vdu + dudv
    [/tex]

    subtract y = uv
    [tex]
    dy = udv + vdu
    [/tex]

    this equates to:
    [tex]
    \frac{dy}{dx} = (-x^2-2x+3)\frac{d}{dx}(x^2+x-2) + (x^2+x-2)\frac{d}{dx}(-x^2-2x+3)
    [/tex]
    [tex]
    \frac{dy}{dx}=(-x^2-2x+3)(-2x-2) + (x^2+x-2)(2x+1)
    [/tex]

    Thats my attempt :P

    EDIT: Corrected
     
    Last edited: Jul 25, 2006
  7. Jul 25, 2006 #6

    VietDao29

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    Homework Helper

    Uhmm, this line is wrong. What you have written in this line is udu + vdv, in stead of udv + vdu.
    Uhmm, you know that:
    [tex]\left( x ^ {\alpha} \right) ' = \alpha x ^ {\alpha - 1}[/tex], right?
    And you should also know that the derivative of a sum, is the sum of derivatives, i.e:
    (u + v + w + k) = u' + v' + w' + k'
    And if k is a constant then:
    [kf(x)]' = k f'(x)
    A polynomial has the form:
    [tex]n_m x ^ m + n_{m - 1} x ^ {m - 1} + ... + n_1 x + n_0[/tex]
    This is a polynomial of degree m.
    The derivative of that polynomial is:
    [tex]\left( n_m x ^ m + n_{m - 1} x ^ {m - 1} + ... + n_1 x + n_0 \right)' = n_m \left( x ^ m \right)' + n_{m - 1} \left( x ^ {m - 1} \right)' + ... + n_1 (x)' + (n_0)'[/tex]
    [tex]= mn_m \left( x ^ {m - 1} \right)' + (m - 1)n_{m - 1} \left( x ^ {m - 2} \right)' + ... + n_1[/tex].
    Uhmm, not very sure what you mean though. Can you type it out?
    --------------------
    The problem ask you to find the derivative of a product. You can use the Product Rule:
    (uv)' = u'v + uv'
    By letting:
    u = 3 - 2x - x2
    and v = x2 + x - 2.
    We have:
    [(3 - 2x - x2) (x2 + x - 2.)]' = (3 - 2x - x2)' (x2 + x - 2.) + (3 - 2x - x2) (x2 + x - 2.)'.
    Can you go from here? :)
     
  8. Jul 25, 2006 #7
    Sorry, that because of my poor latex skills more than my maths
     
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