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Chain rule differentiation

  1. Jun 16, 2014 #1
    The question:
    ImageUploadedByPhysics Forums1402924990.055025.jpg
    This is the solution that was given by my teacher

    Attempt:

    I understand how the work is done until the 3-4 line. Where did the 1-cos2x disappear to in the 4th line?
    I know you can use the outside inside method but try as I might, I can't seem to understand how the final answer was gotten??

    Can someone please tell me what I'm missing here??
     
  2. jcsd
  3. Jun 16, 2014 #2

    D H

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    What is ##(1+\cos(2x)) + (1-\cos(2x))## ?
     
  4. Jun 16, 2014 #3
    The differentiation is done from line 1 to line 2. The rest is just tidying things up a little. The equality from line 3 to line 4 follows simply because

    \begin{equation*}
    2\sin(2x)(1 + \cos(2x)) + 2\sin(2x)(1 - \cos(2x)) =2 \sin(2x)( 1 + \cos(2x) + 1 - \cos(2x)) = 4 \sin(2x).
    \end{equation*}
     
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