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The Slope of the Curve, Derivatives

  1. Aug 26, 2009 #1
    1. Find the derivative f' of f(x)=x^2. Then find the slope of the curve y=f(x) at x=3 and write an equation for the tangent line.


    2. I know that the derivative of x^2 is 2x and I'm guessing that y=x^2, therfore, (3)^2 = 9 which is the value of y. So now the point is (3,9). How do I find the slope? Do I do rise over run which is 9/3 which is 3?

    3. I already know the tangent line equation but the slope is getting me confused. I know slope is y=mx+b But what if this is with derivatives?
     
  2. jcsd
  3. Aug 26, 2009 #2

    nicksauce

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    No. The equation of the line is y = mx + b. m is the slope and b is the y intercept.

    To find the slope of the tangent line, note that the derivative of a function at a point gives the slope of the tangent line. Then you have one point (3,9) and the slope, so you should be able to find the equation of the line.
     
  4. Aug 26, 2009 #3
    The equation of the tangent line is:

    [tex]y-y_0=f'(x_0)(x-x_0)[/tex]

    In this case x0=3 and y0=9

    Regards.
     
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