Chapter 4: Derivatives of Sinusoidal Functions

A

100

Find the derivative:

f(x)=4sinx−3cosx

f′(x)=4cosx+3sinx

200

Differentiate: f(x)=sin(3x).

Using the chain rule: f′(x)=3cos(3x)

300

Differentiate: f(x)=x²sinx.

Using the product rule:

f′(x)=2xsinx+x2cosx

400

Find the second derivative of

f(x)=sin(2x). 

f′(x)=2cos(2x) 

f′′(x)=−4sin(2x) 

500

A particle has position s(t)=5cos(2t)−4sint.

Find the times in 0≤t≤2π

when the particle is momentarily at rest.

v(t)=−10sin(2t)−4cost

Set v(t)=0:    −10sin(2t)−4cost=0

Using sin(2t)=2sintcost, 

        −20sintcost−4cost=0 

        −4cost(5sint+1)=0. 

Thus: cost=0 or sint=−51.

Therefore:  

t=π/2,3π/2,π+sin^-1(1/5),2π−sin^-1(1/5) 

B

100

Find dy/dx if  y=5cosx+2sinx.

dy/dx=−5sinx+2cosx

200

Find the derivative of

y=cos(5x−2).

y′=−5sin(5x−2)

300

Find the derivative of

f(x)=sinx/x. 

Using the quotient rule:

f′(x)=(xcosx−sinx)/x²

400

Find the third derivative of

f(x)=cos(3x). 

f′(x)=−3sin(3x) 

f′′(x)=−9cos(3x) 

f′′′(x)=27sin(3x) 

500

Find the second derivative of

f(x)=xsin(2x). 

First derivative: f′(x)=sin(2x)+2xcos(2x)

Second derivative:

f′′(x)=2cos(2x)+2cos(2x)−4xsin(2x)

Therefore: f′′(x)=4cos(2x)−4xsin(2x) 

C

100

Find the derivative of f(x)=tanx.

f′(x)=sec2x

200

Find f′(x) for f(x)=2sin(4x)+3cos(2x).

f′(x)=8cos(4x)−6sin(2x)

300

Differentiate:

y=(3x²+1).cos(2x). 

y′=6xcos(2x)−2(3x²+1)sin(2x) 

400

Find all values of x in 0≤x≤2π

where f(x)=2cosx+sinx has a horizontal tangent.

f′(x)=−2sinx+cosx

Set equal to zero: cosx=2sinx

                           tanx=1/2

Therefore: x=tan^-1(1/2),π+tan^-1(1/2) 

500

Determine the values of x in 0≤x≤2π 

where f(x)=sinx+cosx has a horizontal tangent.

f′(x)=cosx−sinx

Set equal to zero: cosx=sinx 

                           tanx=1

Thus: x=π/4,5π/4 

D

100

Find f′(0) if f(x)=3sinx+4cosx.

f′(x)=3cosx−4sinx 

f′(0)=3 

3

200

Find the derivative of

y=tan(2x+1). 

Answer:

y′=2sec²(2x+1) 

300

Find f′(x) if f(x)=sin(x²+3x).

f′(x)=(2x+3)cos(x²+3x)

400

Find the equation of the tangent line to

y=sin(2x) at x=π/4. 

Answer:

y′=2cos(2x)

At x=4π:

m=2cosπ/2=0

Also, y=sinπ/2=1.

Therefore: y=1 

500

For f(x)=3sin(2x)−4cos(2x),

find the maximum possible value of f′(x).

f′(x)=6cos(2x)+8sin(2x)

The maximum value of Acosθ+Bsinθ

is  (A²+B²)^1/2.

Therefore: (6²+8²)^1/2= 100^1/2 

                               = 10 

E

100

Find the slope of the tangent to y=sinx at x=π/3.

y′=cosx 

m=cosπ/3 

1/2

200

Find the equation of the tangent line to

f(x)=2sinx at x=π/6. 

f(π/6)=1 

f′(x)=2cosx      f′(π/6)=3^1/2

Therefore,    y−1=3^1/2(x−π/6) 

300

Find the values of x in 0≤x≤2π where

f(x)=sinx

has a horizontal tangent.

Horizontal tangent means

f′(x)=cosx=0.

Thus: x=π/2,3π/2 

400

A particle moves according to

s(t)=4sin(2t)+3t.

Find its velocity and acceleration functions.

Velocity: v(t)=8cos(2t)+3

Acceleration: a(t)=−16sin(2t) 

500

A particle moves along a straight line with position

s(t)=6sint−3cos(2t)

for 0≤t≤2π.

a) Find the velocity function.
b) Find the acceleration function.
c) Determine all times when the particle is at rest.
d) Determine the acceleration at those times.

Velocity v(t)=s′(t) 

            v(t)=6cost+6sin(2t)

Using sin(2t)=2sintcost, 

            v(t)=6cost(1+2sint).

For the particle to be at rest: 

            6cost(1+2sint)=0.

Thus:    cost=0  or sint=−1/2.

On 0≤t≤2π:

The particle is at rest at

t=π/2,7π/6,3π/2,11π/6