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Calculus (chain rule)

  1. Sep 18, 2010 #1
    I am trying to differentiate ((5x-1)^4)((8x^2-5)^-3) but i am stuck at a certain point.....
    Can you please help me fill in the blanks?

    Thank you so much:):)


    Work done:
    1st step: (5x-1)^4 d/dx (8x^2-5)^-3 + (8x^2-5)^-3 d/dx(5x-1)^4

    2nd step: (5x-1)^4 (-3)(8x^2-5)^-4 (16x)(8x^2-5)^-3 + (4)(5x-1)^3 (5)

    3rd step: (I dont know what goes here.....something x) (5x-1)^4 (8x^2-5)^-4 +9(5x-1)^4 (8x^2-5)^-3

    Answer: 4(5x-4)^3 (8x^2-5)^-4 (I dont know what goes here)
     
  2. jcsd
  3. Sep 19, 2010 #2
    f = ((5x-1)^4)((8x^2-5)^-3)
    f' = ((5x-1)^4)'(8x^2-5)^-3) + ((5x-1)^4)((8x^2-5)^-3)'
    f' = (4*5(5x-1)^3)(8x^2-5)^-3) + ((5x-1)^4)(16x*(-3)(8x^2-5)^-4)
    simplify
    f' = 20((5x-1)^3)((8x^2-5)^-3) -48x((5x-1)^4)((8x^2-5)^-4)
    f' = ((5x-1)^3)((8x^2-5)^-4)(-80x^2+48x -100)

    Not sure if this helps
     
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