Derivatives of Trig Functions
Selected Problems
Matthew Staley
September 15, 2011
Derivatives of Trig Functions: Selected Problems
1. Find f ! (x):
(a)
f (x) = 4 cos x + 2 sin x
d
d
(cos x) + 2 (sin x)
dx
dx
= −4 sin x + 2 cos x
f ! (x) = 4
(b)
f (x) = 2 sin2 x = 2 (sin x)2 = 2 sin x sin x
d
(sin x sin x)
dx
d
d
= 2(sin x (sin x) + sin x (sin x))
dx
dx
= 2(sin x cos x + sin x cos x)
= 2(2 sin x cos x)
f ! (x) = 2
= 4 sin x cos x
(c)
f (x) = sec x −
f ! (x) =
√
2 tan x
√ d
d
(sec x) − 2 (tan x)
dx
dx
= sec x tan x −
√
2 sec2 x
1
(d)
f (x) =
5 − cos x
5 + sin x
d
d
(5 − cos x) − (5 − cos x) dx
(5 + sin x)
(5 + sin x) dx
f (x) =
2
(5 + sin x)
!
=
(5 + sin x)(sin x) − (5 − cos x)(cos x)
(5 + sin x)2
=
5 sin x + sin2 x − 5 cos x + cos2 x
(5 + sin x)2
=
(sin2 x + cos2 x) + 5(sin x − cos x)
(5 + sin x)2
=
(e)
1+5(sin x−cos x)
(5+sin x)2
f (x) = sec x tan x
f ! (x) = sec x
d
d
(tan x) + tan x (sec x)
dx
dx
= sec x(sec2 x) + tan x(sec x tan x)
= sec3 x + sec x tan2 x
2
2. Find
(a)
d2 y
:
dx2
y = x cos x
dy
d
d
= cos x (x) + x (cos x)
dx
dx
dx
= cos x − x sin x
d2 y
d
d
=
(cos
x)
−
(x sin x)
dx2
dx
dx
d
d
= − sin x − (sin x (x) + x (sin x))
dx
dx
= − sin x − (sin x + x cos x)
= − sin x − sin x − x cos x
= −2 sin x − x cos x
(b)
y = csc x
dy
d
=
(csc x)
dx
dx
= − csc x cot x
d
d2 y
= − (csc x cot x)
2
dx
dx
d
d
= −(cot x (csc x) + csc x (cot x))
dx
dx
= −(cot x(− csc x cot x) + csc x(− csc2 x))
= cot2 x csc x + csc3 x
3
3. Find the equation of the line tangent to the graph of sin x at x = π/4:
Let f (x) = sin x, then f ( π4 ) = sin ( π4 ) = √12 .
Now f ! (x) = cos x, so f ! ( π4 ) = cos ( π4 ) = √12 .
!
"
We now have a point, π4 , √12 , and a slope,
point-slope formula we obtain our line:
1
1 !
π"
y− √ = √ x−
4
2
2
1
x
π
y−√ = √ − √
2
2 4 2
y=
√x
2
−
π
√
4 2
+
√1 .
2
So plugging this into the
√1
2
4. Show that y = cos x and y = sin x are solutions to the equation y !! + y = 0.
y = cos x
y ! = − sin x
y !! = − cos x
y !! + y = − cos x + cos x = 0.
y = sin x
y ! = cos x
y !! = − sin x
y !! + y = − sin x + sin x = 0.
4
5. A 10-ft ladder leans against a wall at an angle θ. The of top of the ladder is x
feet above the ground. If the bottom of the ladder is pushed towards the wall,
find the rate at which x changes with respect to θ when θ = 60◦ . Express the
answer in units of feet/degree.
10f t !!
x
sin θ =
10
x = 10 sin θ
!
!
!
! θ
dx
d
= 10 (sin θ)
dθ
dθ
= 10 cos θ
#
!π "
dx ##
1
=
10
cos
= 10( ) = 5
#
dθ θ=60◦ = π
3
2
3
$
%
5f t π rad
=
rad 180◦
=
πf t
ft
≈ 0.087
◦
36
deg
5
!
!
x
6. An airplane is flying on a horizontal path at a height of 3800 ft. At what rate
is the distance s between the airplane and the point P changing with respect
to θ when θ = 30◦ ? Express the answer in units of feet/degree.
3800
sin θ =
s
3800
s=
= 3800 csc θ
sin θ
s !!
!
d
ds
= 3800 (csc θ)
dθ
dθ
= −3800 csc θ cot θ
#
!π "
!π "
ds ##
=
−3800
csc
cot
dθ #θ=30◦ = π
6
6
6
√
√
= −3800(2)( 3) = −7600 3
$
%
√ rad π rad
= −7600 3
ft
180◦
√
−380 3π f t
ft
=
≈ −230
9
deg
deg
6
!θ
P!
!
!
!
3800
7. Determine where f is differentiable.
(a) f (x) = sin x
sin x is differentiable for all real values of x because it is continuous for all
real values of x.
(b) f (x) = cot x
cot x =
cos x
sin x
is not continuous whenever sin x = 0, or x = 0, π, 2π, 3π, . . .
So f (x) is differentiable for all x $= nπ, n = 0, ±1, ±2, ±3, . . .
(c) f (x) =
1
1+cos x
This is not continuous whenever
1 + cos x = 0
cos x = −1
x = π, 3π, 5π, . . .
So f (x) is differentiable for all x $= (2n + 1)π, n = 0, ±1, ±2, ±3, . . .
7