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Derivatives using chain rule

  1. Apr 7, 2008 #1

    70f

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    id like some help deriving certain functions using the chain rule
    the way our teacher does it is different from what the textbook says
    he derives the outermost functions before getting to the innermost functions, this is where
    i get confused =(


    for example

    f(x) = sincos(5x)
    i get
    f'(x) = (coscos(5x)) (-sin(5x)) (5)
    = -5coscos(5x)sin(5x)

    im not sure if this is right

    another question:

    f(x) = sin^4(2x) + cos^4(2x)
     
  2. jcsd
  3. Apr 7, 2008 #2

    nicksauce

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    The first looks correct... why not have a stab at the second?
     
  4. Apr 7, 2008 #3

    70f

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    here is what i get

    f(x) = sin^4(2x) + cos^4(2x)

    f'(x) = [ (4sin^3(2x) (cos^4(2x)) (2) ] + [ (4cos^3(2x) (-sin(2x)) (2) ]

    = 8sin^3(2x)cos^4(2x) + -8cos^3(2x)sin(2x)


    another question - this one i get a different answer than my friend

    f(x) = [ ((3x^2 -2)^5) - 1 ] ^5

    i get

    f(x) = (3x^10 - 2^5 - 1)^5
    = 3x^50 - 2^25 - 1^5

    f'(x) = 150x^49

    here is what im working on right now - ill post what i get in a few mins

    f(x) = |x^1/3|
    f(x) = x^2 cos^2(x^2)
    f(x) = [ (2x+5)/(7x-9) ]^2/3
     
  5. Apr 7, 2008 #4

    nicksauce

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    For the first one in your post you're not applying the chain rule correctly.

    d/dx sin(2x)^4 = 4 * sin(2x) ^3 * d/dx (sin2x)

    For your second one

    (3x^10 - 2^5 - 1)^5 is certainly NOT the same as 3x^50 - 2^25 - 1^5
     
  6. Apr 7, 2008 #5

    70f

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    for f(x) = sin^4(2x) + cos^4(2x), i am doing it right im just deriving it wrong? i should derive each part then add together..if so would the answer be


    8sin^3(2x)cos(2x) - 8cos^3(2x)sin(2x)

    for f(x) = [ ((3x^2 -2)^5) - 1 ] ^5
    i get

    f'(x) = (150x)(3x^2 - 2)^24


    another one im sketched on
    f(x) = x^2 cos^2(x^2)

    using product rule i get

    f'(x) = (2x) (cos^2(2x)) + (x^2) (2cos(x^2)(sin(x^2))(2x)
    =2xcos^x(x^2) + 4x^3cos(x^2)sin(x^2)
     
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