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Chain Rule

  1. Oct 17, 2006 #1
    f(t) = (1+tan t)^(1/3) differentiate using chain rule.

    u = 1 + tan t
    y = u^(1/3)

    dy/dt = dy/du x du/dt


    u=1+tan t

    1/3 u^(-2/3) when u = 1 + tan t x sec^(2)t =

    = sec^(2)t/3(1+tan t)^(2/3)

    Did I do this correct??
     
  2. jcsd
  3. Oct 18, 2006 #2
    [tex] f(t) = (1+ \tan t)^{\frac{1}{3}} [/tex].

    [tex] \frac{du}{dt} = sec^{2} t [/tex]

    So it should be [tex] \frac{1}{3}(1+ \tan t)^{-\frac{2}{3}}\sec^{2}t [/tex]
     
  4. Oct 18, 2006 #3
    and How does it apply whenever we have the fractional derivtive operator [tex] D^{q}f(g(x)) [/tex] (1)

    If we wish to calculate the derivative of (1) respect to x, q>0 and real
     
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