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Derivative of sin^2x

  1. Mar 15, 2010 #1
    Hi.

    Please tell me where this reasoning is wrong, because I know it is but I can't see how.
    f(x) = sinx
    f ' (x) = cosx
    f(60) = sin 60
    f ' (60) = cos 60

    but

    g(x) = sin 2x
    g'(x) = 2 cos 2x
    set x = 30
    then: g(30) = sin 60
    g'(x) = 2 cos 60

    but

    f(60) = g(30)
    so f ' (60) = g'(30)
    i.e cos 60 = 2 cos 60
    thus: 1 = 2

    WHY!!!!! HOW!!!!!
     
    Last edited by a moderator: Feb 14, 2013
  2. jcsd
  3. Mar 15, 2010 #2
    Re: Derivative of sin2x

    The fact that two functions cross at a particular point does not mean that their derivative is the same.

    You could do the same thing with much simpler functions eg f(x)=x, g(x)=2x these cross at x=0 but they obviously don't have the same gradient.
     
  4. Mar 15, 2010 #3
    Re: Derivative of sin2x

    Thank you. That is quite obvious, isn't it.
     
  5. Mar 20, 2010 #4
    Re: Derivative of sin2x

    Try this one:

    x=0
    x+1=1
    (x+1)(x+1)=1(x+1)
    x^2+2x+1=x+1
    x^2+x=0
    x(x+1)=0
    x+1=0
    x=-1

    ???
     
  6. Mar 20, 2010 #5
    Re: Derivative of sin2x

    :wink: Ops!
     
  7. Mar 20, 2010 #6
    Re: Derivative of sin2x

    Where did it go wrong algebraically?
     
  8. Mar 20, 2010 #7
    Re: Derivative of sin2x

    I just told you! I boldfaced those parts.
     
  9. Mar 20, 2010 #8

    Char. Limit

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    Gold Member

    Re: Derivative of sin2x

    Actually, you could just as easily say that at...

    x(x+1) = 0

    You could just as easily divide by (x+1) to get...

    x = 0.

    After all, the first equation in this post says that x is EITHER 0 or -1.

    And we know that it's zero.
     
  10. Mar 20, 2010 #9

    rock.freak667

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    Homework Helper

    Re: Derivative of sin2x

    If you do that you might miss out a solution.

    if ab=0, then either a=0 or b=0.

    if you just divide by a, you'll get b=0 only.
     
  11. Mar 20, 2010 #10
    Re: Derivative of sin2x

    ^What he said. Doing that causes a missing solution. By the end, you should get x= -1 or 0. However, -1 is an extraneous solution. Therefore, you're stuck with x=0. Check out http://en.wikipedia.org/wiki/Extraneous_solution if you're confused.
     
  12. Mar 20, 2010 #11

    Char. Limit

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    Gold Member

    Re: Derivative of sin2x

    Yeah, I know that...

    My point was that he missed a solution.
     
  13. Mar 20, 2010 #12

    Redbelly98

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    Re: Derivative of sin2x

    Moderator's note:

    Please keep on topic with the thread. New topics should be started in a new thread.

    (See post #1 if you don't know what the topic of this thread is.)
     
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