Finding the Second Derivative of a Trigonometric Function

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SUMMARY

The discussion focuses on finding the second derivative of the function y' = csc²(θ/2). The user correctly applies the chain rule but initially overlooks the negative sign in the derivative of csc(x), which is -csc(x)cot(x). The simplification of the expression leads to the conclusion that the second derivative can be expressed as csc²(θ/2) * cot(θ/2). The key takeaway is the importance of correctly applying derivative rules, particularly for trigonometric functions.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosecant and cotangent.
  • Knowledge of the chain rule in calculus.
  • Familiarity with half-angle identities in trigonometry.
  • Ability to differentiate basic trigonometric functions.
NEXT STEPS
  • Review the derivative rules for trigonometric functions, focusing on csc(x) and cot(x).
  • Study half-angle identities and their applications in calculus.
  • Practice finding higher-order derivatives of trigonometric functions.
  • Explore the implications of the chain rule in more complex functions.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives of trigonometric functions, and educators looking for examples of applying derivative rules in trigonometry.

cheezeitz
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Homework Statement



y' = csc2([tex]\vartheta[/tex] / 2 )
Find y"



The Attempt at a Solution



So far, i have 2 csc ([tex]\vartheta[/tex]/2) *csc([tex]\vartheta[/tex]/2)cot([tex]\vartheta[/tex]/2) * 1/2

but I'm wondering, how do you take the derivative of a half angle identity?
or does it just simplify down to csc2([tex]\vartheta[/tex] / 2) * cot([tex]\vartheta[/tex]/2)
 
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You don't need to worry about derivatives of half-angle identities here. You're right about how it simplifies, and you used the chain rule properly, except you're missing a minus sign (the derivative of csc(x) is -csc(x)cot(x)).
 

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