Furman University
Derivatives of Trigonometric Functions
Mark R. Woodard
c 2000 Mark.Woodard@Furman.edu
Copyright Last Revision Date: March 14, 2000
Section 1: Derivatives of Trigonometric Functions
2
1. Derivatives of Trigonometric Functions
If f (x) = sin(x), what is f 0 (x)? We have no alternative but to use the definition of derivative. Thus we have
that
sin(x + h) − sin(x)
h→0
h
sin(x) cos(h) − sin(h) cos(x) − sin(x)
= lim
h→0
h
sin(x) cos(h) − sin(x) − sin(h) cos(x)
= lim
h→0
x
sin(x) (cos(h) − 1) − cos(x) sin(h)
= lim
h→0
h
sin(h)
(cos(h) − 1)
= lim sin(x)
− cos(x)
h→0
h
h
sin(h)
cos(h) − 1
− cos(x) lim
.
= sin(x) lim
h→0
h→0
h
h
f 0 (x) = lim
If we denote limh→0 cos(h)−1
by a and limh→0 sin(h)
by b, then we have deduced that the derivative of
h
h
sin(x) is a sin(x) + b cos(x). It remains to compute these two limits.
1.1. Two Important Trigonometric Limits
We will need to use a little geometry to help us compute limh→0
sin(h)
h .
Consider the picture below:
Section 1: Derivatives of Trigonometric Functions
3
Recall that the area of a sector of a circle of radius r and angle h is A = 12 hr2 . Now consider the sector
OSQ. It has area 21 h cos2 (h). On the other hand, triangle OP Q has area 12 cos(h) sin(h), and sector OP R
has area 21 h. From the picture we can tell that even when h is small,
area OSQ ≤ area OP Q ≤ area OP R
so
1
1
1
h cos2 (h) ≤ cos(h) sin(h) ≤ h.
2
2
2
If we multiply through by 2 and divide by cos(h) · h, we see that
cos(h) ≤
1
sin(h)
≤
.
h
cos(h)
Now as h → 0, cos(h) → 1, so this inequality together with the squeeze theorem tell us that limh→0
sin(h)
h
=
1.
Exercise 1. Use the now established fact that limh→0 sin(h)
= 1, to show that limh→0 cos(h)−1
= 0. Hint:
h
h
Try multiplying the numerator and the denominator by the conjugate of the numerator.
1.2. Derivatives of other Trigonometric Functions
d
d
Knowing that dx
sin(x) = cos(x) and dx
cos(x) = − sin(x), and knowing the quotient rule is enough to enable
one to compute the derivatives of the other trigonometric functions, since they are all quotients involving
these two trigonometric functions. For example
d
dx
tan(x)
=
=
=
=
Exercise 2. Compute
d
dx
sec(x),
d
dx
cot(x), and
d sin(x)
dx cos(x)
cos(x) cos(x)−sin(x)(− sin(x))
cos2 (x)
1
cos2 (x)
2
sec (x)
d
dx
csc(x) in a manner similar to the previous example.
Of course, it would be a good idea to practice these new trigonometric derivatives in conjuction with out
familar rules such as the chain rule, quotient rule, product rule, etc..
2 x +1
.
Exercise 3. Compute the derivative with respect to x of f (x) = sin 2x−3
√
√
Exercise 4. Compute g 0 (x) for g(x) = ( x2 − 4) · tan( x2 − 4).
Section 1: Derivatives of Trigonometric Functions
4
Begin Quiz Answer the following questions about trigonometric derivatives.
1. What is the derivative of f (x) =
sec2 (x)
cos(x)
tan(x)
cos(x) ?
cos(x)
sin(2x)
sin2 (x)+1
cos3 (x)
2. What is the derivative of f (x) = tan(cos(x))?
sec2 (cos(x)
tan(− sin(x))
sec2 (cos(x)) · − sin(x)
3. If f (x) = (sin(x2 + cos(x2 )))2 , what is f 0 (x)?
2 sin(x2 + cos(x2 ))
2 sin(x2 + cos(x2 )) · (cos(x2 +
− sin(x2 ))
2 sin(x2 + cos(x2 )) · (cos(x2 +
cos(x2 )) · (2x + − sin(x2 ) · 2x)
p
d
4. Find dx
cot(2x + 1).
p
− csc2 (2x + 1)
1
2 (cot(2x
−1
1
2 ·− csc2 (2x+1)·2
2 (cot(2x+1)
End Quiz
Stewart: Section 3.5, 1 - 23 odd.
Section 3.6, 27 - 37 odd.
+ 1)
−1
2
·1
Solutions to Exercises
5
Solutions to Exercises
Exercise 1.
cos(h) − 1 cos(h) + 1
cos(h) − 1
= lim
·
h→0
h→0
h
h
cos(h) + 1
cos2 (h) − 1
= lim
h→0 h · (cos(h) + 1)
lim
− sin2 (h)
h→0 h · (cos(h) + 1)
sin(h)
− sin(h)
= lim
· lim
h→0
h→0 cos(h) + 1
h
−0
=1·
1+1
=0
= lim
Exercise 1
Exercise 2.
d
dx
sec(x)
=
d
1
dx cos(x)
cos(x)·0−1·(− sin(x))
cos2 (x)
sin(x)
cos2 (x)
=
sec(x) tan(x)
=
=
d
dx
cot(x)
=
=
=
=
d
dx
d cos(x)
dx sin(x)
sin(x)(− sin(x)−cos(x) cos(x))
sin2 (x)
−1
sin2 (x)
2
− csc (x)
csc(x)
=
=
=
d
1
dx sin(x)
sin(x)·0−1·cos(x)
sin2 (x)
− cos(x)
sin2 (x)
= − cot(x) csc(x)
Exercise 2
Exercise 3.
If f (x) = sin
x2 +1
2x−3
, then f 0 (x) = cos
x2 +1
2x−3
·
(2x−3)(2x)−(x2 +1)(2)
.
(2x−3)2
Solutions to Exercises
6
Exercise 3
Exercise 4.
√
√
√
√
√
−1
If g(x) = ( x2 − 4)·tan( x2 − 4), then g 0 (x) = ( x2 − 4)·sec2 ( x2 − 4)· 12 (x2 −4) 2 ·2x+tan( x2 − 4)·
−1
1
2
2 · 2x.
2 (x − 4)
Exercise 4