Differentiating f(t) with Chain Rule

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SUMMARY

The discussion focuses on differentiating the function f(t) = (1 + tan t)^(1/3) using the chain rule. The correct derivative is derived as dy/dt = (1/3)(1 + tan t)^(-2/3) * sec^2(t). The participants confirm the application of the chain rule and explore its implications for fractional derivatives, specifically the operator D^q f(g(x)) for q > 0. The calculations and applications discussed are accurate and relevant for advanced calculus topics.

PREREQUISITES
  • Understanding of the chain rule in calculus
  • Familiarity with trigonometric functions, specifically tangent and secant
  • Knowledge of fractional derivatives and their notation
  • Basic proficiency in differentiation techniques
NEXT STEPS
  • Study the application of the chain rule in more complex functions
  • Learn about fractional calculus and the properties of fractional derivatives
  • Explore the implications of derivatives in real-world applications
  • Investigate advanced differentiation techniques, including implicit differentiation
USEFUL FOR

Students of calculus, mathematicians, and educators seeking to deepen their understanding of differentiation techniques, particularly in relation to trigonometric functions and fractional derivatives.

helpm3pl3ase
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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??
 
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[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]
 
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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