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Differentiation by first principles

  • 06-06-2011 12:54pm
    #1
    Closed Accounts Posts: 2


    Two days from the leaving cert and ive realised i have no idea how to differentiate by first principles, and to make matter worse, I have no notes on it. If anyone could help me out with this id really appreciate it.


Comments

  • Registered Users, Registered Users 2 Posts: 3,038 ✭✭✭sponsoredwalk


    [latex]y_0 \ = \ f(x_0)[/latex]
    [latex]y_0 \ + \ \Delta \ y \ = \ f(x_0 \ + \ \Delta x )[/latex]
    [latex]y_0 \ + \ \Delta y \ - \ y_0 \ = \ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)[/latex]
    [latex] \Delta y \ = \ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)[/latex]
    [latex] \frac{ \Delta y}{ \Delta x}} \ = \ \frac{ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)}{ \Delta x}[/latex]
    [latex] \frac{dy}{dx} \ = \ \lim_{\Delta x \rightarrow 0} \ \frac{ f(x_0 \ + \ \Delta x ) \ - \ f(x_0)}{ \Delta x}[/latex]

    So:

    [latex]f(x) \ = \ 2x[/latex]
    [latex]f(x \ + \ \Delta x) \ = \ 2(x \ + \ \Delta x)[/latex]
    [latex]f(x \ + \ \Delta x) \ - \ f(x) = \ 2(x \ + \ \Delta x) \ - \ 2x[/latex]
    .
    .
    .

    Try [latex]f(x) \ = \ \frac{x}{x \ - \ 1}[/latex].


  • Closed Accounts Posts: 2 Stuffnmusic


    thanks, i owe you one :)


  • Registered Users, Registered Users 2 Posts: 5,637 ✭✭✭TheBody


    Check out the following video. It should have everything you need.

    http://patrickjmt.com/derivatives-using-the-definition-to-find-a-derivative/


  • Registered Users, Registered Users 2 Posts: 73 ✭✭jody24


    differentiate 10-2x with respect to x from first principles


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  • Closed Accounts Posts: 43 Fish Finger Pie


    10


  • Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    10
    Err, newp.

    It's almost three years since the OP did their Leaving Cert. Whether they managed to down with calculus or not is fairly moot by now, I'd say.

    IBTL.


  • Closed Accounts Posts: 43 Fish Finger Pie


    So what's the answer then? <snip>


  • Registered Users, Registered Users 2 Posts: 7,157 ✭✭✭srsly78


    The "answer" is to show the method step by step.


  • Closed Accounts Posts: 43 Fish Finger Pie


    y = 10 - 2x

    As x tends to 0, y tends to 10

    Therefore, dy/dx = 10.

    Right?


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  • Registered Users, Registered Users 2 Posts: 7,157 ✭✭✭srsly78


    Right final answer, wrong method. See post #2. It doesn't say x tends to 0, it says delta x tends to 0.

    edit: oops not even the right final answer!


  • Closed Accounts Posts: 43 Fish Finger Pie


    y = 10 - 2x

    dy = 10 - 2dx

    dy/dx = 10/dx - 2 = 10


  • Registered Users, Registered Users 2 Posts: 3,038 ✭✭✭sponsoredwalk


    y = 10 - 2x

    dy = 10 - 2dx

    dy/dx = 10/dx - 2 = 10

    When you write y = 10 - 2x what you really mean is y(x) = 10 - 2x.
    Then y(x + Δx) = 10 - 2(x + Δx).
    So y(x + Δx) - y(x) = [10 - 2(x + Δx)] - [10 - 2x] = (10 - 10) - 2x + 2x - 2Δx = - 2Δx.
    Then [y(x + Δx) - y(x)]/Δx = - 2.
    Taking a limit as Δx ---> 0 gives dy/dx = - 2.


  • Closed Accounts Posts: 43 Fish Finger Pie


    Where'd the 10 go to?


  • Registered Users, Registered Users 2 Posts: 7,157 ✭✭✭srsly78


    10 - 10 = 0


  • Moderators, Science, Health & Environment Moderators Posts: 1,852 Mod ✭✭✭✭Michael Collins


    This is going nowhere. Locked.


This discussion has been closed.
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