Section3.3 Derivatives of trigonometric functions 1.

Section3.3

Derivatives of trigonometric functions

1. f (x) = 1− 3 sin x

⇒ f⁰(x) = (1)⁰− 3(sin x)⁰ =−3 cos x

2. f (x) = x sin x

⇒ f⁰(x) = (x)⁰sin x + x(sin x)⁰ = sin x + x cos x

9. y = _{2−tan x}^x

⇒ ^dy_dx = (2−tan x)(x)⁰−x(2−tan x)⁰

(2−tan x)² = ⁽²−tan x)(1)−x(− sec²x)

(2−tan x)² = ²−tan x+x sec²x (2−tan x)²

22. y = e^xcos x, (0, 1)

dy

dx = e^xcos x + e^x(− sin x) = e^xcos x− e^xsin x

dy

dx|x=0 = e⁰cos 0− e⁰sin 0 = 1− 0 = 1

The slope of the tangent line at (0, 1) is 1. So, the equation of the tangent line at (0, 1) is y− 1 = 1(x − 0) or y = x + 1.

35. (a). x(t) = 8 sin t

The equation of velocity v(t) = x⁰(t) = 8 cos t.

The equation of acceleration a(t) = x⁰⁰(t) =−8 sin t.

(b). The mass at time t = ^2π₃ has position x(^2π₃ ) = 8 sin(^2π₃ ) = 8(^√₂³) = 4√ 3, ve- locity v(^2π₃ ) = 8 cos(^2π₃ ) = 8(⁻¹₂ ) =−4, and acceleration a(^2π₃ ) =−8 sin(^2π₃ ) =

−8(^√₂³) =−4√

3. Since v(^2π₃ ) < 0, the particle is moving to the left.

37. From the figure 1: 3-3-12, we can see that sin θ = x/6⇔ x = 6 sin θ. We want to find the rate of change of x with respect to θ, that is ^dx_dθ. Taking the derivative of x = 6 sin θ, we get ^dx_dθ = 6 cos θ. So, when θ = ^π₃, ^dx_dθ = 6 cos(^π₃) = 6(¹₂) = 3 m/rd

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Figure 1: 3-3-12

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