Chain Rule

  • #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??
 

Answers and Replies

  • #2
1,235
1
[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]
 
  • #3
131
0
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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