Does the Derivative of the Integral of Tan(x) Consider Cos(x) Sign Changes?

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SUMMARY

The discussion centers on the derivative of the integral of the tangent function, specifically ∫tan(x) dx = -ln|cos(x)| + C. The derivative f'(x) is correctly identified as -sin(x)/|cos(x)|. It is established that when taking the derivative, one does not restrict f'(x) to intervals where cos(x) > 0, as the properties of logarithmic differentiation apply universally, regardless of the sign of cos(x).

PREREQUISITES
  • Understanding of integral calculus, specifically integration of trigonometric functions.
  • Familiarity with logarithmic differentiation and its properties.
  • Knowledge of the behavior of trigonometric functions, particularly sine and cosine.
  • Basic differentiation rules and their application to composite functions.
NEXT STEPS
  • Study the properties of logarithmic differentiation in depth.
  • Explore the implications of sign changes in trigonometric functions on derivatives.
  • Learn about the integral and derivative relationships of other trigonometric functions.
  • Investigate advanced topics in calculus, such as the Fundamental Theorem of Calculus.
USEFUL FOR

Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of trigonometric integrals and derivatives.

MathewsMD
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∫tan(x) dx = -ln lcos(x)l + C = f(x)

So is f'(x) = -sin(x)/lcos(x)l ? When taking the derivative, do we only consider f'(x) to only exist on the intervals where cos(x) > 0 instead?
 
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MathewsMD said:
∫tan(x) dx = -ln lcos(x)l + C = f(x)

So is f'(x) = -sin(x)/lcos(x)l ? When taking the derivative, do we only consider f'(x) to only exist on the intervals where cos(x) > 0 instead?

No, the derivative of log(|y(x)|) is the derivative of log(y(x)) if y(x)>0 and it's the derivative of log(-y(x)) if y(x)<0. Can you show that they are both (1/y(x))*y'(x). Not (1/|y(x)|)*y'(x)?
 

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