Lesson 3-4 Derivatives of Trigonometric Functions

Objectives Find the derivatives of trigonometric functions

Vocabulary none new

Trig Differentiation Rules d ---- (sin x) = cos x Sin dx d ---- ( cos x) = -sin xCos dx Rest of them can be done from these two and the Quotient rule!

Rest of Trig Differentiation Rules d ---- [tan x] = sec² x dx d ---- (cot x) = -csc² x dx d ---- [sec x] = (sec x) (tan x) dx d ---- (csc x) = -(csc x) (cot x) dx

Example 1 1.y = sin x – cos x 2.f(x) = (tan x) / (x + 1) Find the derivatives of the following: y’(x) = cos x – (-sin x) = cos x + sin x (x + 1) (sec² x) – (1) (tan x) y’(x) = (x + 1)²

Example 2 3.y = sin (π/4) 4.y = x³ sin x Find the derivatives of the following: y’(x) = 0 y’(x) = 3x² (sin x) + x³ (cos x)

Example 3 5.y = x² + 2x cos x 6.y = x / (sec x + 1) Find the derivatives of the following: y’(x) = 2x + 2x (-sin x) + (2) (cos x) (sec x + 1) (1) – (sec x tan x) (x) y’(x) = (sec x + 1)²

Example 4 7.y = x / (cot x) 8. y = (csc x) / e x Find the derivatives of the following: (cot x) (1) – (-csc² x) (x) y’(x) = (cot x)² (e x ) (-csc x cot x) – (e x ) (csc x) y’(x) = (e x )²

Example 5 sin x 9. lim x sin 5x 10. lim x Find the derivatives of the following: x→0 sin x = (1/7) lim = (1/7) (1) = 1/7 x x→0 sin u = lim = 1 u u→0 letting u = 5x

Example 6 A particle moves along a line so that at any time t>0 its position is given by x(t) = 2πt + cos(2πt). Find the velocity at time t. Find the speed (|v|) at t = ½. What are the values of t for which the particle is at rest? v(t) = x’(t) = 2π – 2π (sin 2πt) Speed = |v(1/2)| = |x’(t)| = |2π – 2π (sin π)| = 2π When v(t) = 0 = 2π – 2π (sin 2πt) 2π (sin 2πt) = 2π  sin 2πt = 1  t = ¼, 5/4, 9/4, 13/4, etc

Summary & Homework Summary: –Use trig rules for finding derivatives – d(sin x) = cos x – d(cos x) = -sin x Homework: –pg 216 – 217: 1-3, 6, 9-11, 18, 29, 41