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Derivative of trigonometric functions

  1. Sep 8, 2008 #1
    y = sin( 4 x ) cos( 3 x )

    f(x) = sin4x
    g(x) = cos3x
    f'(x) = cos4x
    g'(x) = -sin3x

    And by using the product rule, i'll get:

    cos4x(cos3x) - sin4x(sin3x)

    Is the answer correct or can be simplify again??
     
  2. jcsd
  3. Sep 8, 2008 #2
    Check the f'(x) and g'(x). Are you missing something?
     
  4. Sep 8, 2008 #3
    sin4x can be written in this way (4(sin x)) or not?
     
    Last edited: Sep 8, 2008
  5. Sep 8, 2008 #4
  6. Sep 8, 2008 #5
    Is it

    f'(x)=4cos(4x)

    ??
     
    Last edited: Sep 8, 2008
  7. Sep 8, 2008 #6
    Yes :smile:
    And the other?
     
  8. Sep 8, 2008 #7
    so the answer would be

    4cos4x(cos3x) -3 sin4x(sin3x)

    Correct??
     
  9. Sep 8, 2008 #8
    Yes.
     
  10. Sep 8, 2008 #9
    can my last answer be simplify one more step?
     
  11. Sep 8, 2008 #10
    Not any that I can think of. It can be re-arranged. Do you have a target answer?
     
  12. Sep 8, 2008 #11
    Try to draw the graphs for y= 4 sinx and y=sin4x for x= 0,30,45,60,90 degrees etc. and see for yourself.
     
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