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15. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
47
15. Derivatives of Trigonometric Functions
In this section, we show how to compute the derivatives of trigonometric functions. First, we compute (sin x) . This will be a somewhat
lengthy procedure, due to the fact that this is the first trigonometric
function we will differentiate and we will have to apply new methods.
However, once we know the derivatives of sin x and cos x, it will be
much simpler to deduce the derivatives of other trigonometric functions, since those functions can be obtained from sin and cos, and then
the various differentiation rules can be used.
Theorem 16. We have (sin x) = cos x.
Proof. Recall the identity sin(a + b) = sin a cos b + sin b cos a. We
have
sin(x + h) − sin x
(sin x) = lim
h→0
h
sin x cos h + sin h cos x − sin x
= lim
h→0
h
(cos h) − 1 sin h cos x
+
= lim sin x
h→0
h
h
(cos h) − 1
sin h
= sin x lim
+ cos x lim
.
h→0
h→0 h
h
Note that as, h approaches 0, we certainly have limh→0 sin x = sin x
and limh→0 cos x = cos x, since these functions do not even depend on
h.
There remains the task of computing the two nontrivial limits
(cos h) − 1
sin h
and lim
.
h→0
h
h
We will carry out this task in two lemmas.
lim
h→0
Lemma 1. We have
sin h
= 1.
h→0 h
Proof. Let us consider a circle with unit radius and a regular ngon whose center is at the center O of the circle and whose n vertices
are all on the unit circle. Then the area of the circle is π, and the area
of the n-gon is n · 12 · sin α, where α = 2π/n is the angle AOB, with A
and B being adjacent vertices of our n-gon.
Considering just 1/n of both the circle and the n-gon, we see that
the area of the triangle AOB is (sin α)/2, and the area of 1/n of the
lim
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3. RULES OF DIFFERENTIATION
circle bordered by the lines AO, BO, and the arc AB is π · α/(2π) =
α/2. So the ratio of the two areas is
(sin α)/2
sin α
.
=
α/2
α
On the other hand, as n gets larger and larger, α gets smaller and
smaller, while the area of the n-gon gets closer and closer to the area
of the circle. Hence, their ratio, sin α/α, will get arbitrarily close to 1
and stay arbitrarily close to 1.
2
Lemma 2. The equality
(3.4)
lim
h→0
(cos h) − 1
=0
h
holds.
Proof. We will manipulate the expression ((cos h) − 1)/h so that
we can use the result of Lemma 1. First, we multiply both the numerator and the denominator by cos h + 1 to get
(cos2 h) − 1
− sin2 h
(cos h) − 1
=
=
.
h
h(1 + cos h)
h(1 + cos h)
Therefore, we have
(cos h) − 1
sin2 h
= − lim
h→0
h→0 h(1 + cos h)
h
sin h
sin h
·
= − lim
h→0 h
1 + cos h
sin h
sin h
= − lim
· lim
h→0 h
h→0 1 + cos h
= (−1) · 0 = 0.
lim
2
We can now finish the proof of Theorem 16. At the end of the first
displayed chain of equations in that proof, we saw that
(cos h) − 1
sin h
+ cos x lim
.
h→0
h→0 h
h
The previous two lemmas showed that, on the right-hand side, the first
limit is 0 and the second limit is 1, so (sin x) = cos x as claimed. 2
(sin x) = sin x lim
The following theorem can be proved by very similar methods.
Theorem 17. The equality (cos x) = − sin x holds.
16. THE CHAIN RULE
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You are asked to prove this theorem in Exercise 15.1(1).
Now that we have the derivatives of sin and cos, the derivatives
of other trigonometric functions can be obtained by simply using the
quotient rule. The next theorem shows an example of this.
Theorem 18. We have (tan x) = sec2 x.
Proof. Note that tan x = sin x/cos x, so we can apply the quotient
rule. This leads to
sin x
(tan x) =
cos x
cos x · (sin x) − sin x(cos x)
=
cos2 x
cosx + sin2 x
=
cos2 x
1
=
cos2 x
2
= sec2 x.
The derivatives of the other three trigonometric functions are given
in the exercises.
15.1. Exercises.
(1)
(2)
(3)
(4)
(5)
(6)
Prove that (cos x) = − sin x.
Prove that (cot x) = − csc2 x.
Prove that (csc x) = − csc x · cot x.
Prove that (sec x) = sec x tan x.
Let h(x) = ex cos x. Find h (x).
Let h(x) = ex / sin x. Find h (x).
16. The Chain Rule
16.1. The Derivative of the Composition of Two Functions. In previous
sections, we learned how to compute the derivative of the sum, difference, product, and quotient of two functions. We still do not know
how to compute the derivative
of the composition of functions, such as
√
h(x) = sin(3x), t(x) = x2 + 1, or r(x) = e4x . In this section, we will
learn a rule, called the chain rule, that applies in these situations.
Theorem 19 (Chain Rule). Let h(x) = f (g(x)), where g is differentiable at x and f is differentiable at g(x). Then h is differentiable at
x, and we have
h (x) = f (g(x))g (x).