Skip to Content
OpenStax Logo
Calculus Volume 1

3.5 Derivatives of Trigonometric Functions

Calculus Volume 13.5 Derivatives of Trigonometric Functions
Buy book
  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 3.5.1. Find the derivatives of the sine and cosine function.
  • 3.5.2. Find the derivatives of the standard trigonometric functions.
  • 3.5.3. Calculate the higher-order derivatives of the sine and cosine.

One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and cosine functions will enable us to find the velocity and acceleration of simple harmonic motion.

Derivatives of the Sine and Cosine Functions

We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its derivative. Recall that for a function f(x),f(x),

f(x)=limh0f(x+h)f(x)h.f(x)=limh0f(x+h)f(x)h.

Consequently, for values of hh very close to 0, f(x)f(x+h)f(x)h.f(x)f(x+h)f(x)h. We see that by using h=0.01,h=0.01,

ddx(sinx)sin(x+0.01)sinx0.01ddx(sinx)sin(x+0.01)sinx0.01

By setting D(x)=sin(x+0.01)sinx0.01D(x)=sin(x+0.01)sinx0.01 and using a graphing utility, we can get a graph of an approximation to the derivative of sinxsinx (Figure 3.25).

The function D(x) = (sin(x + 0.01) − sin x)/0.01 is graphed. It looks a lot like a cosine curve.
Figure 3.25 The graph of the function D(x)D(x) looks a lot like a cosine curve.

Upon inspection, the graph of D(x)D(x) appears to be very close to the graph of the cosine function. Indeed, we will show that

ddx(sinx)=cosx.ddx(sinx)=cosx.

If we were to follow the same steps to approximate the derivative of the cosine function, we would find that

ddx(cosx)=sinx.ddx(cosx)=sinx.
Theorem 3.8

The Derivatives of sin x and cos x

The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.

ddx(sinx)=cosxddx(sinx)=cosx
3.11
ddx(cosx)=sinxddx(cosx)=sinx
3.12

Proof

Because the proofs for ddx(sinx)=cosxddx(sinx)=cosx and ddx(cosx)=sinxddx(cosx)=sinx use similar techniques, we provide only the proof for ddx(sinx)=cosx.ddx(sinx)=cosx. Before beginning, recall two important trigonometric limits we learned in Introduction to Limits:

limh0sinhh=1andlimh0cosh1h=0.limh0sinhh=1andlimh0cosh1h=0.

The graphs of y=(sinh)hy=(sinh)h and y=(cosh1)hy=(cosh1)h are shown in Figure 3.26.

The function y = (sin h)/h and y = (cos h – 1)/h are graphed. They both have discontinuities on the y-axis.
Figure 3.26 These graphs show two important limits needed to establish the derivative formulas for the sine and cosine functions.

We also recall the following trigonometric identity for the sine of the sum of two angles:

sin(x+h)=sinxcosh+cosxsinh.sin(x+h)=sinxcosh+cosxsinh.

Now that we have gathered all the necessary equations and identities, we proceed with the proof.

ddxsinx=limh0sin(x+h)sinxhApply the definitionof the derivative.=limh0sinxcosh+cosxsinhsinxhUse trig identity for the sine of the sum of two angles.=limh0(sinxcoshsinxh+cosxsinhh)Regroup.=limh0(sinx(cosh1h)+cosx(sinhh))Factor outsinxandcosx. =sinx·0+cosx·1Apply trig limit formulas.=cosxSimplify.ddxsinx=limh0sin(x+h)sinxhApply the definitionof the derivative.=limh0sinxcosh+cosxsinhsinxhUse trig identity for the sine of the sum of two angles.=limh0(sinxcoshsinxh+cosxsinhh)Regroup.=limh0(sinx(cosh1h)+cosx(sinhh))Factor outsinxandcosx. =sinx·0+cosx·1Apply trig limit formulas.=cosxSimplify.

Figure 3.27 shows the relationship between the graph of f(x)=sinxf(x)=sinx and its derivative f(x)=cosx.f(x)=cosx. Notice that at the points where f(x)=sinxf(x)=sinx has a horizontal tangent, its derivative f(x)=cosxf(x)=cosx takes on the value zero. We also see that where f(x)=sinxf(x)=sinx is increasing, f(x)=cosx>0f(x)=cosx>0 and where f(x)=sinxf(x)=sinx is decreasing, f(x)=cosx<0.f(x)=cosx<0.

The functions f(x) = sin x and f’(x) = cos x are graphed. It is apparent that when f(x) has a maximum or a minimum that f’(x) = 0.
Figure 3.27 Where f(x)f(x) has a maximum or a minimum, f(x)=0f(x)=0 that is, f(x)=0f(x)=0 where f(x)f(x) has a horizontal tangent. These points are noted with dots on the graphs.

Example 3.39

Differentiating a Function Containing sin x

Find the derivative of f(x)=5x3sinx.f(x)=5x3sinx.

Solution

Using the product rule, we have

f(x)=ddx(5x3)·sinx+ddx(sinx)·5x3=15x2·sinx+cosx·5x3.f(x)=ddx(5x3)·sinx+ddx(sinx)·5x3=15x2·sinx+cosx·5x3.

After simplifying, we obtain

f(x)=15x2sinx+5x3cosx.f(x)=15x2sinx+5x3cosx.

Checkpoint 3.25

Find the derivative of f(x)=sinxcosx.f(x)=sinxcosx.

Example 3.40

Finding the Derivative of a Function Containing cos x

Find the derivative of g(x)=cosx4x2.g(x)=cosx4x2.

Solution

By applying the quotient rule, we have

g(x)=(sinx)4x28x(cosx)(4x2)2.g(x)=(sinx)4x28x(cosx)(4x2)2.

Simplifying, we obtain

g(x)=−4x2sinx8xcosx16x4=xsinx2cosx4x3.g(x)=−4x2sinx8xcosx16x4=xsinx2cosx4x3.

Checkpoint 3.26

Find the derivative of f(x)=xcosx.f(x)=xcosx.

Example 3.41

An Application to Velocity

A particle moves along a coordinate axis in such a way that its position at time tt is given by s(t)=2sintts(t)=2sintt for 0t2π.0t2π. At what times is the particle at rest?

Solution

To determine when the particle is at rest, set s(t)=v(t)=0.s(t)=v(t)=0. Begin by finding s(t).s(t). We obtain

s(t)=2cost1,s(t)=2cost1,

so we must solve

2cost1=0for0t2π.2cost1=0for0t2π.

The solutions to this equation are t=π3t=π3 and t=5π3.t=5π3. Thus the particle is at rest at times t=π3t=π3 and t=5π3.t=5π3.

Checkpoint 3.27

A particle moves along a coordinate axis. Its position at time tt is given by s(t)=3t+2costs(t)=3t+2cost for 0t2π.0t2π. At what times is the particle at rest?

Derivatives of Other Trigonometric Functions

Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use the quotient rule to find formulas for their derivatives.

Example 3.42

The Derivative of the Tangent Function

Find the derivative of f(x)=tanx.f(x)=tanx.

Solution

Start by expressing tanxtanx as the quotient of sinxsinx and cosx:cosx:

f(x)=tanx=sinxcosx.f(x)=tanx=sinxcosx.

Now apply the quotient rule to obtain

f(x)=cosxcosx(sinx)sinx(cosx)2.f(x)=cosxcosx(sinx)sinx(cosx)2.

Simplifying, we obtain

f(x)=cos2x+sin2xcos2x.f(x)=cos2x+sin2xcos2x.

Recognizing that cos2x+sin2x=1,cos2x+sin2x=1, by the Pythagorean theorem, we now have

f(x)=1cos2x.f(x)=1cos2x.

Finally, use the identity secx=1cosxsecx=1cosx to obtain

f(x)=sec2x.f(x)=sec2x.
Checkpoint 3.28

Find the derivative of f(x)=cotx.f(x)=cotx.

The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these formulas in the following theorem.

Theorem 3.9

Derivatives of tanx,cotx,secx,tanx,cotx,secx, and cscxcscx

The derivatives of the remaining trigonometric functions are as follows:

ddx(tanx)=sec2xddx(tanx)=sec2x
3.13
ddx(cotx)=csc2xddx(cotx)=csc2x
3.14
ddx(secx)=secxtanxddx(secx)=secxtanx
3.15
ddx(cscx)=cscxcotx.ddx(cscx)=cscxcotx.
3.16

Example 3.43

Finding the Equation of a Tangent Line

Find the equation of a line tangent to the graph of f(x)=cotxf(x)=cotx at x=π4.x=π4.

Solution

To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute

f(π4)=cotπ4=1.f(π4)=cotπ4=1.

Thus the tangent line passes through the point (π4,1).(π4,1). Next, find the slope by finding the derivative of f(x)=cotxf(x)=cotx and evaluating it at π4:π4:

f(x)=csc2xandf(π4)=csc2(π4)=−2.f(x)=csc2xandf(π4)=csc2(π4)=−2.

Using the point-slope equation of the line, we obtain

y1=−2(xπ4)y1=−2(xπ4)

or equivalently,

y=−2x+1+π2.y=−2x+1+π2.

Example 3.44

Finding the Derivative of Trigonometric Functions

Find the derivative of f(x)=cscx+xtanx.f(x)=cscx+xtanx.

Solution

To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find

f(x)=ddx(cscx)+ddx(xtanx).f(x)=ddx(cscx)+ddx(xtanx).

In the first term, ddx(cscx)=cscxcotx,ddx(cscx)=cscxcotx, and by applying the product rule to the second term we obtain

ddx(xtanx)=(1)(tanx)+(sec2x)(x).ddx(xtanx)=(1)(tanx)+(sec2x)(x).

Therefore, we have

f(x)=cscxcotx+tanx+xsec2x.f(x)=cscxcotx+tanx+xsec2x.
Checkpoint 3.29

Find the derivative of f(x)=2tanx3cotx.f(x)=2tanx3cotx.

Checkpoint 3.30

Find the slope of the line tangent to the graph of f(x)=tanxf(x)=tanx at x=π6.x=π6.

Higher-Order Derivatives

The higher-order derivatives of sinxsinx and cosxcosx follow a repeating pattern. By following the pattern, we can find any higher-order derivative of sinxsinx and cosx.cosx.

Example 3.45

Finding Higher-Order Derivatives of y=sinxy=sinx

Find the first four derivatives of y=sinx.y=sinx.

Solution

Each step in the chain is straightforward:

y=sinxdydx=cosxd2ydx2=sinxd3ydx3=cosxd4ydx4=sinx.y=sinxdydx=cosxd2ydx2=sinxd3ydx3=cosxd4ydx4=sinx.

Analysis

Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so

d4dx4(sinx)=d8dx8(sinx)=d12dx12(sinx)==d4ndx4n(sinx)=sinxd5dx5(sinx)=d9dx9(sinx)=d13dx13(sinx)==d4n+1dx4n+1(sinx)=cosx.d4dx4(sinx)=d8dx8(sinx)=d12dx12(sinx)==d4ndx4n(sinx)=sinxd5dx5(sinx)=d9dx9(sinx)=d13dx13(sinx)==d4n+1dx4n+1(sinx)=cosx.
Checkpoint 3.31

For y=cosx,y=cosx, find d4ydx4.d4ydx4.

Example 3.46

Using the Pattern for Higher-Order Derivatives of y=sinxy=sinx

Find d74dx74(sinx).d74dx74(sinx).

Solution

We can see right away that for the 74th derivative of sinx,74=4(18)+2,sinx,74=4(18)+2, so

d74dx74(sinx)=d72+2dx72+2(sinx)=d2dx2(sinx)=sinx.d74dx74(sinx)=d72+2dx72+2(sinx)=d2dx2(sinx)=sinx.
Checkpoint 3.32

For y=sinx,y=sinx, find d59dx59(sinx).d59dx59(sinx).

Example 3.47

An Application to Acceleration

A particle moves along a coordinate axis in such a way that its position at time tt is given by s(t)=2sint.s(t)=2sint. Find v(π/4)v(π/4) and a(π/4).a(π/4). Compare these values and decide whether the particle is speeding up or slowing down.

Solution

First find v(t)=s(t):v(t)=s(t):

v(t)=s(t)=cost.v(t)=s(t)=cost.

Thus,

v(π4)=12.v(π4)=12.

Next, find a(t)=v(t).a(t)=v(t). Thus, a(t)=v(t)=sinta(t)=v(t)=sint and we have

a(π4)=12.a(π4)=12.

Since v(π4)=12<0v(π4)=12<0 and a(π4)=12>0,a(π4)=12>0, we see that velocity and acceleration are acting in opposite directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling. Consequently, the particle is slowing down.

Checkpoint 3.33

A block attached to a spring is moving vertically. Its position at time tt is given by s(t)=2sint.s(t)=2sint. Find v(5π6)v(5π6) and a(5π6).a(5π6). Compare these values and decide whether the block is speeding up or slowing down.

Section 3.5 Exercises

For the following exercises, find dydxdydx for the given functions.

175.

y=x2secx+1y=x2secx+1

176.

y=3cscx+5xy=3cscx+5x

177.

y=x2cotxy=x2cotx

178.

y=xx3sinxy=xx3sinx

179.

y=secxxy=secxx

180.

y=sinxtanxy=sinxtanx

181.

y=(x+cosx)(1sinx)y=(x+cosx)(1sinx)

182.

y=tanx1secxy=tanx1secx

183.

y=1cotx1+cotxy=1cotx1+cotx

184.

y=cosx(1+cscx)y=cosx(1+cscx)

For the following exercises, find the equation of the tangent line to each of the given functions at the indicated values of x.x. Then use a calculator to graph both the function and the tangent line to ensure the equation for the tangent line is correct.

185.

[T] f(x)=sinx,x=0f(x)=sinx,x=0

186.

[T] f(x)=cscx,x=π2f(x)=cscx,x=π2

187.

[T] f(x)=1+cosx,x=3π2f(x)=1+cosx,x=3π2

188.

[T] f(x)=secx,x=π4f(x)=secx,x=π4

189.

[T] f(x)=x2tanx,x=0f(x)=x2tanx,x=0

190.

[T] f(x)=5cotx,x=π4f(x)=5cotx,x=π4

For the following exercises, find d2ydx2d2ydx2 for the given functions.

191.

y=xsinxcosxy=xsinxcosx

192.

y=sinxcosxy=sinxcosx

193.

y=x12sinxy=x12sinx

194.

y=1x+tanxy=1x+tanx

195.

y=2cscxy=2cscx

196.

y=sec2xy=sec2x

197.

Find all xx values on the graph of f(x)=−3sinxcosxf(x)=−3sinxcosx where the tangent line is horizontal.

198.

Find all xx values on the graph of f(x)=x2cosxf(x)=x2cosx for 0<x<2π0<x<2π where the tangent line has slope 2.

199.

Let f(x)=cotx.f(x)=cotx. Determine the points on the graph of ff for 0<x<2π0<x<2π where the tangent line(s) is (are) parallel to the line y=−2x.y=−2x.

200.

[T] A mass on a spring bounces up and down in simple harmonic motion, modeled by the function s(t)=−6costs(t)=−6cost where ss is measured in inches and tt is measured in seconds. Find the rate at which the spring is oscillating at t=5t=5 s.

201.

Let the position of a swinging pendulum in simple harmonic motion be given by s(t)=acost+bsints(t)=acost+bsint where aa and bb are constants, tt measures time in seconds, and ss measures position in centimeters. If the position is 0 cm and the velocity is 3 cm/s when t=0t=0, find the values of aa and bb.

202.

After a diver jumps off a diving board, the edge of the board oscillates with position given by s(t)=−5costs(t)=−5cost cm at tt seconds after the jump.

  1. Sketch one period of the position function for t0.t0.
  2. Find the velocity function.
  3. Sketch one period of the velocity function for t0.t0.
  4. Determine the times when the velocity is 0 over one period.
  5. Find the acceleration function.
  6. Sketch one period of the acceleration function for t0.t0.
203.

The number of hamburgers sold at a fast-food restaurant in Pasadena, California, is given by y=10+5sinxy=10+5sinx where yy is the number of hamburgers sold and xx represents the number of hours after the restaurant opened at 11 a.m. until 11 p.m., when the store closes. Find yy and determine the intervals where the number of burgers being sold is increasing.

204.

[T] The amount of rainfall per month in Phoenix, Arizona, can be approximated by y(t)=0.5+0.3cost,y(t)=0.5+0.3cost, where tt is months since January. Find yy and use a calculator to determine the intervals where the amount of rain falling is decreasing.

For the following exercises, use the quotient rule to derive the given equations.

205.

ddx(cotx)=csc2xddx(cotx)=csc2x

206.

ddx(secx)=secxtanxddx(secx)=secxtanx

207.

ddx(cscx)=cscxcotxddx(cscx)=cscxcotx

208.

Use the definition of derivative and the identity

cos(x+h)=cosxcoshsinxsinhcos(x+h)=cosxcoshsinxsinh to prove that d(cosx)dx=sinx.d(cosx)dx=sinx.

For the following exercises, find the requested higher-order derivative for the given functions.

209.

d3ydx3d3ydx3 of y=3cosxy=3cosx

210.

d2ydx2d2ydx2 of y=3sinx+x2cosxy=3sinx+x2cosx

211.

d4ydx4d4ydx4 of y=5cosxy=5cosx

212.

d2ydx2d2ydx2 of y=secx+cotxy=secx+cotx

213.

d3ydx3d3ydx3 of y=x10secxy=x10secx

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/calculus-volume-1/pages/1-introduction
Citation information

© Mar 30, 2016 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.