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Derivative of a line

  1. Nov 2, 2004 #1
    Hi again! Time for one more of my newbie questions.
    I'm reading "Elementary Calculus: An Approach Using Infinitesimals
    " http://www.math.wisc.edu/~keisler/calc.html and can't get a theorem on a lines derivative. It goes like this:
    [tex]f(x)=kx+b \Rightarrow \frac{dy}{dx} = f'(x) = k[/tex]

    That doesn't make seens to me because the definition of a tangent line is
    g(x)=f'(x)(x-a)+b, there (a, b) is the point of the tangent.

    For instance, let's say f(x)=2x. The f'(x)=2, using the above theorem. And then the tangent for the point (2, 2) ought to be l(x)=2(x-2)+2=2x-4+2=2x-2. But it's parallell to f(x)=2x!! :surprised What am I doing wrong, please give me a hint!
     
  2. jcsd
  3. Nov 2, 2004 #2

    arildno

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    Eeh, the point is (1,2), not (2,2), so:
    l(x)=2(x-1)+2=2x
     
  4. Nov 2, 2004 #3
    Ohh. I took a point OUTSIDE the line. I feel so dumb...
     
  5. Nov 2, 2004 #4
    Thanks for the support. By the way, many from Scandinavia here, isn't. The reson to care is that I'm a Swede. Anyway, maybe enought OT talk now.
     
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