Using Trig Identities to see if derrivatives are equal

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SUMMARY

The discussion focuses on finding two derivatives of the function y = sec(x) * cot(x) and determining their equality. The first derivative, calculated using the product rule, is dy/dx = -sec(x) * cot²(x), while the second derivative, obtained through simplification, is dy/dx = -csc(x) * cot(x). The user seeks confirmation on whether these two expressions are equivalent, which is validated through the identities sec(x) = 1/cos(x) and cot(x) = cos(x)/sin(x).

PREREQUISITES
  • Understanding of trigonometric identities, specifically secant and cotangent functions.
  • Familiarity with derivative rules, particularly the product rule.
  • Basic algebraic manipulation skills to simplify trigonometric expressions.
  • Knowledge of calculus concepts related to derivatives and their applications.
NEXT STEPS
  • Study trigonometric identities in depth, focusing on secant and cosecant functions.
  • Practice applying the product rule in various calculus problems.
  • Explore algebraic techniques for simplifying trigonometric expressions.
  • Learn about the implications of derivative equality in calculus.
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and trigonometric functions, as well as educators looking for examples of derivative verification using trigonometric identities.

keykar
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Homework Statement


for homework we have to find 2 different dirrivatives of the same problem (one may be incorrect) and then tell if they are equal

2. The attempt at a solution
original
y=sec(x)*cot(x)

the two derrivatives (i know these are correct becase i have compared with others in the class)

using product rule first
dy/dx=-sec(x)*cot^2(x)

using simplification first
dy/dx=-csc(x)*cot(x)

this is my question...is there any way u can find that -secx(cotx)^2=-cscxcotx ??
 
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Sure. You should know that sec(x) = 1/cos(x) and that cot(x) = cos(x)/sin(x). Using these definitions we have that sec(x) * cot2(x) = (1/cos(x)) * (cos2(x)/sin2(x)) = . . .
 
Thx jgens! Ur a life saver
 

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