CHAPTER 21 Derivatives of Trig Functions

CHAPTER 21

Derivatives of Trig Functions

n Part 3 we have introduced the idea of a derivative of a function, which

I we defined in terms of a limit. Then we began the task of finding rules

that compute derivatives withouht liimits. Here is our list of rules so far.

Constant function rule:

Dx c

= 0

hi

Identity function rule:

= Dx x 1

Power rule:

hi

n=

? n1

Dx x nx

hi

Exponential rule:

x= x Dx e e

hi

Constant multiple rule:

Dx c f (x)

=0 c f (x)

Sum-di erence rule:

h

i

?

= 0 ?0

Dx f (x) g(x) f (x) g (x)

Product rule:

h

i

=0

+

0

Dx f (x)g(x) f (x)g(x) f (x)g (x)

Quotient rule:

h

i

0

?

0

f (x) = f (x)g(x) f (x)g (x)

Dx

(2

g(x)

g x)

In this chapter we will expand this list by adding six new rules for the

derivatives of the six trigonometric functions:

h

i

h

i

h

i

h

i

h

i

h

i

Dx sin(x) Dx tan(x) Dx sec(x) Dx cos(x) Dx csc(x) Dx cot(x)

This will require a few ingredients. First, we will need the addition formulas for sine and cosine (Equations 3.12 and 3.13 on page 46):

sin(? + ?) = sin(?) cos(?) + cos(?) sin(?) cos(? + ?) = cos(?) cos(?) ? sin(?) sin(?)

2+ 2 = Recall also from Chapter 3 the fundamental identity sin (x) cos (x) 1.

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Derivatives of Trig Functions

And we will need these limits from Theorem 10.2 (Chapter 10, page 152):

?

cos(h) 1 =

li!m

0

h0 h

sin(h) =

and li!m

1.

h0 h

= Let's start by computing the derivative of f (x) sin(x).

0=

+? f (x h) f (x)

f (x) li!m

h0

h

+?

= sin(x h) sin(x)

li!m

h0

h

+

?

= sin(x) cos(h) cos(x) sin(h) sin(x)

li!m

h0

h

?

+

= sin(x) cos(h) sin(x) cos(x) sin(h)

li!m

h0

?

?h

?+

= sin(x) cos(h) 1 cos(x) sin(h)

li!m h 0?

?

h? ?

=

sin(x) cos(h) 1 + cos(x) sin(h)

li!m

h0

?

h ?

h

=

cos(h) 1 +

sin(h)

li!m sin(x)

cos(x)

h0

h

h

(Definition 16.1) =

( f (x) sin(x)) (addition formula for sin)

(regroup) (factor out sin(x)) (break up fraction)

?

=

? cos(h) 1 +

? sin(h)

li!m sin(x) li!m

li!m cos(x) li!m

h0

h0 h

h0

h0 h

(limit laws)

=

?+

?

sin(x) 0 cos(x) 1

(Theorem 10.2)

= cos(x).

Therefore the derivative of sin(x) is cos(x). This is our latest derivative rule.

h

i

Rule 9

Dx sin(x)

= cos(x)

This rule makes sense when we compare the graph of sin(x) with its

?

derivative cos(x). The tangent to sin(x) has slope 0 at integer multiples of ,

2

= and these are exactly the places that cos(x) 0. And notice that where the

tangent to sin(x) has positive slope, cos(x) is positive; where the tangent to

sin(x) has negative slope, cos(x) is negative. As the derivative of sin(x), cos(x)

equals the slope of the tangent to sin(x) at (x, sin(x)).

y

=

=

y sin(x) y cos(x)

??

??

?

?

3?

? 2

5?

? 3

x

2

2

2

2

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?

?

So what is the d?erivat?ive of cos(x)?

Since Dx

sin(x)

= cos(x), you might

= first guess that Dx cos(x) sin(x). But this is not quite right because for

0 < x < ? the tangents to cos(x) have neg? ative ?slope, while sin(x) is positive.

=? However, the graphs below suggest Dx cos(x) sin(x).

y= y cos(x)

??

??

?

?

3?

? 2

5?

? 3

x

2

2

2

2

=? y sin(x)

In fact this turns out to be exactly right. This chapter's Exercise 19 asks

you to adapt the computation on the previous page to get the following rule.

h

i

Rule 10

Dx cos(x)

=? sin(x)

We now have derivative rules for sin and cos. Next, let's compute the

derivative of tan(x). We will use the rules we just derived in conjunction

with the quotient rule and familiar identities.

h

i

=

sin(x)

Dx tan(x) Dx

cos(x)

h

i

h

i

? = Dx sin(x) cos(x) sin(x)Dx cos(x)

2

cos (x)

?

?

?

?

= cos(x) cos(x) sin(x) sin(x)

2+ 2 = cos (x) sin (x)

2

cos (x)

2

cos (x)

?

2

=1= 1

=2 sec (x)

2

cos (x) cos(x)

h

i

We have a new rule: Rule 11

Dx tan(x)

=2 sec (x)

Exercise 20 asks you to do a similar computation to show that

h

i

Rule 12

=? 2

Dx cot(x)

csc (x)

These two latest formulas fit the shapes of the graphs of tan and cot as suggested by Figure 21.1.

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= y tan(x) y

Derivatives of Trig Functions

= y cot(x) y

??

?

2?

3? x ??

?

?

?x

2

3

Figure

21.1.

Any

tangent

line

to

the

graph

of

= y

tan(x)

has

positive

slope.

0= 2 Indeed the slope of the tangent at x is the positive number y sec (x). Any

= tangent line to the graph of y cot(x) has negative slope; the slope of the

0=? 2 tangent at x is the negative number y csc (x).

There are just two more trig functions to consider: sec and csc. We have

hi

h

i

h

i

?

??

=

1 = Dx 1 cos(x) 1 Dx cos(x)

Dx sec(x) Dx

2

cos(x)

cos (x)

?

?

?

? ??

= 0 cos(x) 1 sin(x)

2

cos (x)

= sin(x) = 1 ? sin(x) = sec(x) tan(x).

2

cos (x) cos(x) cos(x)

h

i

This is our latest rule. Rule 13

Dx sec(x)

= sec(x) tan(x)

This chapter's Exercise 21 asks you to do a similar computation to prove

h

i

Rule 14

=?

Dx csc(x)

csc(x) cot(x) .

We now have derivative rules for all six trig functions, which was this chapter's goal. Here is a summary of what we've discovered.

Derivatives of Trig Functions

h

i

= Dx sin(x) cos(x)

h

i

=? Dx cos(x) sin(x)

h

i

=2 Dx tan(x) sec (x)

h

i

=? 2 Dx cot(x) csc (x)

h

i

= Dx sec(x) sec(x) tan(x)

h

i

=? Dx csc(x) csc(x) cot(x)

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Example 21.1

= sin(x) Find the derivative of y 2 + .

x1

This is a quotient, so we use the quotient rule combined with our new rule

for the derivative of sin.

h

i?

?

2+ ?

hi 2+

sin(x) Dx 2 +

x1

= Dx sin(x)

x1 ?

sin(x)Dx x ?

2+ 2

x1

1

?

?

2+ ? = cos(x) x? 1 ? sin(x)2x

2+ 2 x1

Example 21.2

2+ 3

+?

Find the derivative of x x tan(x) .

This is the sum of a power, a product and a constant, so we begin with the

sum-di erence rule, breaking the problem into three separate derivatives,

then using applicable rules.

h

i

hi

h

i

hi

2+ 3

+? =

2+

3

+?

Dx x x tan(x)

Dx x Dx x tan(x) Dx

hi

h

i

=+

3

+3

+

2x Dx x tan(x) x Dx tan(x) 0

|

{z

}

product rule

= +2

+3 2

2x 3x tan(x) x sec (x)

With practice you will quickly reach the point where you will do such a

problem in your head, in one step. (You may already be there.)

w

Example 21.3

= e sec(w)

dz

If z

, find the derivative .

w

dw

This is a quotient, so our first step is to apply the quotient rule.

?

?

??

w

? +w

?

dz = Dw e sec(w) w e sec(w) Dw w

dw ?

2

w

?

w

This now involves Dw e sec(w) , and that requires the product rule.

? ??

?

??

w

+w

? +w

?

= Dw e sec(w) e Dw sec(w) w e sec(w) 1

2

?

w?

w

+w

? +w

= e sec(w) e sec(w) tan(w) w e sec(w)

2

w

?

?

w

+

+

= e sec(w) w w tan(w) 1

2

w

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