Lesson 3 4 Derivatives of Trigonometric Functions Objectives

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

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

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

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

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

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. v(t) = x’(t) = 2π – 2π (sin 2πt) Find the speed (|v|) at t = ½. Speed = |v(1/2)| = |x’(t)| = |2π – 2π (sin π)| = 2π What are the values of t for which the particle is at rest? When v(t) = 0 = 2π – 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