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Using differentials in the equation of tangent line to the curve x^2 at point 3

  1. Jan 14, 2010 #1
    To understand differentials better, I'm trying to use differentials dy and dx in the equation of the tangent line to the curve x^2 at point 3.

    Here is the equation of the tangent line to the curve x^2 at point 3:

    [tex]y=f'(3)(x-3)+f(3)=2(3)(x-3)+9=6(x-3)+9[/tex]

    But since we are dealing with the tangent, why can't we write the above equation as:

    [tex]y=\frac{dy}{dx}(x-3)+dy[/tex]

    Since [tex]\frac{dy}{dx}=6[/tex] from which [tex]dy=6dx[/tex], we can rewrite the above as:

    [tex]6(x-3)+6dx=6(x-3)+6(3)=6(x-3)+18[/tex]

    Why this equation and the first equation aren't the same?

    How else can the equation of tangent line be written in terms of dys and dxs?
     
    Last edited: Jan 14, 2010
  2. jcsd
  3. Jan 15, 2010 #2

    HallsofIvy

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    The dy/dx in place of f' is fine but dy is NOT 9!

    Again, it is because the added "9" is NOT dy and so not "6dx". You are not distinguishing between "dy/dx" and "[itex]\Delta y/\Delta x[/itex]"
    [tex]\frac{dy}{dx}= \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}[/tex]
    You can't just ignore the limit.

    Well, if y= (dy/dx)(x- a)+ y(a) then dx(y- y(a))= dy(x-a). Does that help? It is at least more "symmetric" than the usual form.
     
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