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3 different derivatives?

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the tangent line and the normal line to the curve y=(1+2x)^2 at the point (1,9).


    2. Relevant equations
    f'(x)= [f(x+h) - f(x)] / h ->plug in x=1
    f'(x)= [f(x+h) - f(x)] / h ->plug in (x+h)
    Power Rule
    Power Rule (after factoring out the polynomial)




    3. The attempt at a solution
    I tried to solve it 4 different ways using the above 4 different approaches:

    Plugging in x=1 for the derivative formula yields 12
    Plugging in (x+h) for the derivative formula yields 4+8x
    Using the power rule for (1+2x)^2 = 2(1+2x) = 2+4x (simplified version of 4+8x)
    Using the power rule after factoring out the polynomial is 1+4x+4x^2 = 4+8x


    Regarding the first attempt, I thought I could plug in the x-coordinate to get the tangent equation. Since when is that wrong?
     
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    You have to use both the power rule and the chain rule for the derivative of (1+2x)^2. You get 2*(1+2x)*d(1+2x)/dx=4*(1+2x).
     
  4. Sep 20, 2007 #3
    Well 4*(1+2x) = 4+8x = 2+4x which is what I got.
     
  5. Sep 20, 2007 #4

    Dick

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    Since when is 4+8x=2+4x?
     
  6. Sep 20, 2007 #5

    D H

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    fk378, I rearranged your OP a bit to make a point.

    Dick confirmed this is correct.
    You forgot the factor of 2 when you take the derivative of 1+2x. With this factor of 2 you get the same result as above.
    Obviously this is not a global result; it is only valid at x=1. One question: What is 4+8x evaluated at x=1?
     
  7. Sep 20, 2007 #6
    4+8x simplified = 2+4x
     
  8. Sep 20, 2007 #7

    But isn't the question asking what the slope is at x=1? So why wouldn't plugging in work?
     
  9. Sep 20, 2007 #8

    Dick

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    Yes. And 2 simplified is 1. So 2=1.
     
  10. Sep 20, 2007 #9

    D H

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    I ask again, what is 4+8x evaluated at x=1?
     
  11. Sep 20, 2007 #10
    Sorry D H,
    yes it equals 1. But if it's asking for the equation of the tangent line, what do I regard 12 as?

    Dick,
    2 in the above equations is a common factor that can be factored out. 2=1 has no such common factor.
     
  12. Sep 20, 2007 #11

    D H

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    How in the world is 4+8x evaluated at x=1 equal to 1? (Answer: 4+8x evaluated at x=1 is 12.) How in the world is 4+8x = 2+4x? (Answer: Its not.)

    You appear have some very basic understanding issues here. I think you need some face-to-face time with your teacher.
     
  13. Sep 20, 2007 #12
    Oh I did mean 12. Sorry I was just thinking about the problem...
    Is it not possible to simplify the equation? Does that just apply to derivatives?
     
  14. Sep 20, 2007 #13

    D H

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    You can say 2*(2+4x) = 4+8x. You cannot say 2+4x=4+8x. They are different equations. You can't just throw that factor of two into the ozone layer.
     
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