MA4001 Engineering Mathematics 1 Lecture 14 Derivatives of ...

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

.

Differentiation Critical points

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

Differentiation Critical points

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

Graphs

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

Graphs

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)

Differentiation Critical points

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).

Differentiation Critical points

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.

Differentiation Critical points

Example

y = x3 ⇒ y ′ = 3x2 = 0 at x = 0

Thus x = 0 is a critical point. But it is neither a maximum nor aminimum.