Derivatives of Trigonometric Functions

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SUMMARY

The discussion focuses on finding the critical numbers of the function y = cos x - sin x within the interval -π ≤ x ≤ π. The derivative is correctly calculated as dy/dx = -(sin x + cos x). Setting the derivative to zero leads to the equation sin x + cos x = 0, which simplifies to tan x = -1, indicating the critical points occur where the tangent function equals -1.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine and cosine.
  • Knowledge of derivatives and how to compute them.
  • Familiarity with critical points in calculus.
  • Basic understanding of the tangent function and its properties.
NEXT STEPS
  • Study the unit circle to understand the values of sine and cosine at key angles.
  • Learn how to solve trigonometric equations, particularly those involving tangent.
  • Explore the concept of critical points and their significance in function analysis.
  • Review the application of the first derivative test to determine local maxima and minima.
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding the behavior of trigonometric functions and their derivatives.

DMac
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[SOLVED] Derivatives of Trigonometric Functions

I need to find the critical numbers of this function:
y = cos x - sin x where -pi <= x <= pi

I found the derivative as:

dy/dx = -(sin x + cos x)

But when I equate dy/dx to zero, I get:

sin x + cos x = 0...where do I go from here?
 
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Well when does sinx = -cosx?
 
Ha, lolz I can't believe I didn't think of tan x = -1. (It's getting late, and I've only had 5 hours of sleep these past two nights.) Thanks for the help. =D
 

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