Differentiation Critical points
MA4001 Engineering Mathematics 1Lecture 14
Derivatives of Trigonometric FunctionsCritical Points
Dr. Sarah Mitchell
Autumn 2014
Differentiation Critical points
An important limit
To calculate the limits of basic trigonometric functions we need
to know the limit limx→0
sin xx
.
From the graph it looks like the limit might be 1 (if the function iscontinuous).
Differentiation Critical points
limx→0
sin xx
As the function is even the two one-sided limits must be the
same, so we just need to check limx→0+
sin xx
.
As x → 0 we can assume that 0 < x <π
2and think of x as an
angle in the first quadrant.
Differentiation Critical points
In this graph the point P has coordinates cos x , sin x and T hascoordinates (1, tan x).The area of triangle OAP < the area of the sector OAP < thearea of the triangle OAT .
Differentiation Critical points
Using the formula that the area of a triangle is12× base ×
perpendicular height:
Area of triangle OAP =12× 1 × sin x .
Area of triangle OAT =12× 1 × tan x .
Also the area of a sector of a circle with angle θ is12θ× radius2.
(e.g., angle 2π, radius r , area=πr2)
Thus area of sector OAP is12× x × 12 =
x2
.
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We thus have the bounds:
12
sin x <x2<
12
tan x
Thus
1 <x
sin x<
1cos x
or, inverting this:
cos x <sin x
x< 1
Thus by the squeeze theorem, since
1 = limx→0
cos x 6 limx→0
sin xx
6 limx→0
1 = 1
we have limx→0
sin xx
= 1
Differentiation Critical points
Other limits
We can then obtain
limx→0
tan xx
= limx→0
sin xx cos x
= limx→0
sin xx
limx→0
1cos x
= 1
And also: limx→0
1 − cos xx2 = lim
x→0
2 sin2 x/2x2
Let t =x2
, so that x = 2t and limt→0
≡ limx→0
Differentiation Critical points
Thus
limx→0
1 − cos xx2 = lim
t→0
2 sin2 t(2t)2 =
12
limt→0
(
sin tt
)2
=12
(
limt→0
sin tt
)2
=12
Remark
These limits imply that for small values of x (in radians):
sin x ≈ x ; tan x ≈ x and1 − cos x
x2 ≈12
or cos ≈ 1 −x2
2
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Derivative of sin x
ddx
sin x = limh→0
sin(x + h) − sin xh
Using the formula sin a − sin b = 2 sina − b
2· cos
a + b2
we thus
have
ddx
sin x = limh→0
2 sin h2 cos(x + h
2)
h
= limh/2→0
sin h2
h2
limh→0
cos(x +h2)
= 1 · cos x
= cos x
Differentiation Critical points
Derivative of cos x
ddx
cos x =ddx
sin(π
2− x
)
=ddx
sin u where u =π
2− x
=ddu
sin u ·dudx
by the chain rule
= (cos u) · (−1)
= − cos(π
2− x
)
= − sin x
Differentiation Critical points
We can generalise these derivatives using the chain rule
ddx
f (kx + c) = f ′(kx + c) · k
and haveddx
sin(kx + c) = k cos(kx + c)
andddx
cos(kx + c) = −k sin(kx + c)
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Derivative of tan x and cot x
ddx
tan x =ddx
sin xcos x
=(cos x)(sin x) ′ − (sin x)(cos x) ′
cos2 x
=(cos x)(cos x) − (sin x)(− sin x)
cos2 x
=cos2 x + sin2 x
cos2 x
=1
cos2 x= sec2 x
Similarly one can show thatddx
cot x = − csc2 x .
Differentiation Critical points
Derivative of sec x and csc x
ddx
sec x =ddx
1cos x
= −(cos x) ′
cos2 x(by the rule for reciprocals)
=sin x
cos2 x= (sec x)(tan x)
Similarly one can show thatddx
csc x = −(csc x)(cot x).
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Summary
f (x) f ′(x)sin x cos x
cos x − sin x
tan x sec2 x
cot x − csc2 x
sec x (sec x)(tan x)
csc x −(csc x)(cot x)
Differentiation Critical points
Example:ddx
(
x2 sin√
x)
ddx
(
x2 sin√
x)
= (x2) ′ sin√
x + x2(sin√
x) ′ (by the product rule)
= 2x sin√
x + x2 cos√
xddx
√x (by the chain rule)
= 2x sin√
x + x2 cos√
x1
2√
x
= 2x sin√
x +12
x√
x cos√
x
Differentiation Critical points
Example
Find the tangent line to y = tanπx4
at the point (1,1).
If the slope of the tangent line is m then the equation of thetangent line is
y−y0 = m(x−x0) that is y−tanπ
4= m(x−1) i.e., y−1 = m(x−1)
The slope m is
ddx
tanπx4
∣
∣
∣
∣
x=1=
π
4sec2 πx
4
∣
∣
∣
∣
x=1=
π
4sec2 π
4=
π4
cos2 π4=
π
2
Thus the tangent line is
y − 1 =π
2(x − 1)
Differentiation Critical points
Critical points
Definition
A critical point of the function f (x) is a point where f ′(x) = 0.
Theorem
If f (x) is differentiable in (a,b) and achieves a maximum orminimum at c ∈ (a,b), then f ′(c) = 0, i.e., c is a critical point.
Differentiation Critical points
Proof.
If f (x) has a maximum at c then
f (x) − f (c) 6 0
for all x ∈ (a,b).
Thus f ′(c) = limh→0+
f (c + h) − f (c)h
6 0, since h > 0, and
f ′(c) = limh→0−
f (c + h) − f (c)h
> 0, since h 6 0.
Thus f ′(c) = 0. The case of a minimum is similar.
Remark
f ′(c) = 0 does not imply that there is a maximum or minimumat x = c.