1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    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

    User Avatar
    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

    User Avatar
    Gold Member

    Re: Derivative of sin2x

    Yeah, I know that...

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

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Derivative of sin^2x
  1. Integral of sin(2x)dx (Replies: 2)

Loading...