engineering analysis 1 : differentiation · we use implicit differentiation: treat y as an unknown...

Limits Basic Ideas and Definitions Rules of Differentiation Parametric and Implicit Differentiation Higher Derivatives Optimum Values L’Hopital’s Rule

Engineering Analysis 1 : Differentiation

Dr. Paul D. Ledger, Dr Rob Daniels, Dr Igor Sazonov

[email protected]

College of Engineering, Swansea University, UK

PDL, RD, IS (CoE) WS 2016 1/ 40


Outline

1 Limits

2 Basic Ideas and Definitions

3 Rules of Differentiation

4 Parametric and Implicit Differentiation

5 Higher Derivatives

6 Optimum Values

7 L’Hopital’s Rule

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Introduction to limits

A limit looks at what happens when a function approaches a certain value.

This can be useful if a function can’t be evaluated at a point, but instead we can seewhat happens when we get closer and closer.

Example

What happens as x approaches 1 for f (x) =x2 − 1x− 1

SolutionFor x = 1 then f (x) = 0

0 is undefined and we say it is indeterminate.

But we can investigate what happens as x approaches 1 from belowx 0.5 0.9 0.99 0.999 0.9999 0.99999 · · ·

f (x) 1.5 1.9 1.99 1.999 1.9999 1.99999 · · ·or from above:

x 1.5 1.1 1.01 1.001 1.0001 1.00001 · · ·f (x) 2.5 2.1 2.01 2.001 2.0001 2.00001 · · ·

Thus we see that f (x) approaches 2 as x approaches 1.

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Introduction to limits

More formally we say the limit of f (x) as x approaches 1 from below is 2 or:

limx→1−

f (x) = 2

and the limit of f (x) as x approaches 1 from above is 2 or:

limx→1+

f (x) = 2

Since limx→1− f (x) = limx→1+ f (x) then we say that limit limx→1 f (x) = 2 exists.ExampleIf f (x) = H(x− 1) what is limx→1− f (x) and limx→1+ f (x)? Does limx→1 f (x) exist?SolutionBy considering the definition of the Heaviside unit step function we have

limx→1−

f (x) = limx→1−

H(x− 1) = limx→0−

H(x) = 0

limx→1+

f (x) = limx→1+

H(x− 1) = limx→0+

H(x) = 1

as limx→1− f (x) 6= limx→1+ f (x) then limx→1 f (x) does not exist in this case.

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Rules for calculating limits

If b and c are constants thenlimx→c

b = b

Furthermore if limx→c

f (x) = L and limx→c

g(x) = M and k is a constant

limx→c

kf (x) = k · limx→c

f (x) = kL

limx→c

(f (x) + g(x)) = limx→c

f (x) + limx→c

g(x) = L + M

limx→c

f (x)g(x) = limx→c

f (x) limx→c

g(x) = LM

limx→c

f (x)

g(x)=

limx→c f (x)

limx→c g(x)=

LM

if M 6= 0

Example

Find the limit of f (x) = limx→2

4x3 + 5x + 7(x− 4)(x + 10)

if it exists.

SolutionThe limit of the numerator is lim

x→24x3 + 5x + 7 = 49.

The limit of the denominator islimx→2

(x− 4)(x + 10) = limx→2

(x− 4) limx→2

(x + 10) = (2− 4)(2 + 10) = −24 6= 0 and so

limx→2

4x3 + 5x + 7(x− 4)(x + 10)

= −4924

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Outline

1 Limits





6 Optimum Values


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Drawing Tangents

1

x

y

A

B

θ2

x

y

A

B

θθ

x

y

A

Notice that as the point B moves towards point A then the slope of the chord AB tendsto the slope of the tangent to the curve at point A.

The gradient of the curve at A is the gradient of the tangent at that point. Thegradient at A is also the instantaneous rate of change of the curve at A.

The gradient of the curve at the point A is tan θ where θ is the angle the tangent linemakes with the x axis.

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Drawing Tangents

ExampleFor the curve shown below, state whether the tangent to the curve at points A, B, C, Dand E has positive, negative or zero gradient.

E

y

x

A

BC

D

SolutionA negative, B zero, C positive, D zero and E negative

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Basic Ideas and Definitions

The (automated) process of finding gradients of tangents to graphs in their solution iscalled differentiation and it measures the rate of change of the value of the functionswith respect to its argument.

The gradient of the graph is called the derivative of the function.

The derivative of a function f (x) at the point x is formally defined as

lim∆x→0

f (x + ∆x)− f (x)

∆x= lim

∆x→0

∆f∆x

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Notation for the Derivative

Two kinds of notation are commonly used for representing the derivative. The first usesa composite symbol

dfdx

or df/dx or Df

The second notation uses a primef ′(x)

this means that

dfdx

= f ′(x) = Df = lim∆x→0

∆f∆x

= lim∆x→0

f (x + ∆x)− f (x)

∆x

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Graphical Interpretation

Consider a small change in the dependent variable denoted by ∆x. The correspondingchange in f (x) is given by ∆f

Chord PQ

xx x+ x

x

f(x)

P

Q

∆

∆

∆f

Tangent at P

The slope of the line PQ is∆f∆x

=f (x + ∆x)− f (x)

∆xNow in the limit as ∆x→ 0 the point P→ Q and the segment becomes the tangent tothe curve at P, whose slope is given by the derivative

dfdx

= lim∆x→0

∆f∆x

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Differentiable Functions

A function f (x) can only be differentiated at points where it has a well defined tangent

Examples of functions with points without well defined tangents are:

-1 -0.5 0 0.5 1 1.5 2 2.5 3

x

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

f(x)

-10 -5 0 5 10

x

-3

-2

-1

0

1

2

3

4

f(x)

The function on the left is differentiable at all points except x = 1 and the function onright at all points except x = 0.

For practical purposes it is sufficient to interpret a differentiable function as one havinga smooth continuous graph with no sharp corners. Engineers commonly refer to suchfunctions as being well–behaved.

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Outline

1 Limits





6 Optimum Values


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Standard Derivatives

f (x) f ′(x)

xn where n is real nxn−1

sin x cos xcos x − sin xex ex

Each of the standard derivatives can be obtained by applying the definition

f ′(x) =dfdx

= lim∆x→0

f (x + ∆x)− f (x)

∆x

(and series expansions, see EG190).ExampleBy applying first principles, show that f ′(x) = 2x if f (x) = x2.SolutionBy applying the definition we obtain

f ′(x) =dfdx

= lim∆x→0

(x + ∆x)2 − x2

∆x= lim

∆x→0

x2 + (∆x)2 + 2x∆x− x2

∆x= 2x

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Rules of Differentiation

Rule 1 (scaler multiplication rule)If y = f (x) and k is a constant then

ddx

(ky) = kdydx

= kf ′(x)

Rule 2 (sum rule)If u = f (x) and v = g(x) then

ddx

(u + v) =dudx

+dvdx

= f ′(x) + g′(x)

Rule 3 (product rule)If u = f (x) and v = g(x) then

ddx

(uv) = udvdx

+ vdudx

= f (x)g′(x) + g(x)f ′(x)

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Rules of Differentiation Continued

Rule 4 (quotient rule)If u = f (x) and v = g(x) then

ddx

(uv

)=

v(

dudx

)− u

(dvdx

)v2

=g(x)f ′(x)− f (x)g′(x)

[g(x)]2

Rule 5 (composite–function or chain rule)If z = g(x) and y = f (z) then

dydx

=dydz

dzdx

= f ′(z)g′(x)

Rule 6 (inverse–function rule)If y = f−1(x) then x = f (y) and

dydx

=1

dx/dy=

1f ′(y)

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Examples

QuestionFind the derivative of 2x2

SolutionWe use the scalar multiplication rule

ddx

(2x2) = 2ddx

(x2) = 4x

QuestionFind the derivative of sin x + xSolutionWe use the sum rule

ddx

(x + sin x) =ddx

(x) +ddx

(sin x) = 1 + cos x

QuestionFind the derivative of x3

cos xSolutionWe use the quotient rule

ddx

(x3

cos x

)=

cos x ddx (x3)− x3 d

dx (cos x)

cos2 x=

3x2 cos x + x3 sin xcos2 x

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Examples

QuestionFind the derivative of cos sin xSolutionWe use the chain rule with z = sin x

ddx

(cos sin x) =ddz

(cos z)ddx

(sin x) = − sin z cos x = − sin sin x cos x

QuestionFind the derivative of ln x.SolutionIf y = ln x then x = ey so that

dxdy

= ey

Then, from the inverse–function rule

dydx

=1ey

=1x

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Table of Derivatives

f (x) f ′(x)

xn (n ∈ R) nxn−1

sin x cos xcos x − sin xtan x sec2 xsec x sec x tan xcosec x −cosec x cot xcot x −cosec2 xex ex

ln x (x ∈ R+) 1x

sin−1 x (x ∈ [−1, 1]) 1√1−x2

cos−1 x (x ∈ [−1, 1]) − 1√1−x2

tan−1 x 1(1+x2)

sinh x cosh xcosh x sinh xtanh x 1

cosh2 x= 1− tanh2 x

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More Examples

ExampleShow by using only the derivatives for the elementary functions, show that f ′(t) = sec2 tif f (t) = tan t.Solution

Note that f (t) =g(t)h(t)

with g(t) = sin t and h(t) = cos t thus

g′(t) = cos t h′(t) = − sin t

Then by the quotient rule

f ′(t) =ddt

(g(t)h(t)

)=

h(t)g′(t)− g(t)h′(t)(h(t))2

=cos2 t + sin2 t

cos2 t=

1cos2 t

= sec2 t

where the identity 1 = cost + sin2 t was applied.

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More Examples

ExampleShow by using only the derivatives for the elementary functions, show thatf ′(p) = −cosec p cot p if f (p) = cosec p.SolutionWe realise that

f (p) =g(p)

h(p)=

1sin p

g(p) = 1, h(p) = sin p

so by the quotient rule

f ′(p) =h(p)g′(p)− g(p)h′(p)

(h(p))2= −

cos psin2 p

= −1

sin p tan p= −cosec p cot p

by using the function definitions.

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More Examples

ExampleShow by using only the derivatives for the elementary functions, show thatf ′(w) = 1√

1−w2if f (w) = sin−1 w.

SolutionSet y = f (w) = sin−1 w then w = f−1(y) = siny. Application of the inverse function rulegives

dwdy

= cos ydfdw

=dydw

=1

dw/dy=

1cos y

By using the idenitiy cos2 y + sin2 y = 1 it follows that

dfdw

=1

cos y=

1√1− sin2 y

=1

√1− x2

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Outline

1 Limits





6 Optimum Values


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Parametric Differentiation

A curve may not always be specified in the explicit form y = f (x).

Often curve are expressed parametrically as x = h(t) and y = g(t) where t is commonindependent variable. An example is the curve defining a unit circle x = cos t , y = sin twhere 0 ≤ t ≤ 2π.

The derivativedydx

is obtained from a combination of the chain rule and inverse function:

dydx

=dydt

dtdx

=dydt

1dxdt

=

dydtdxdt

=g′(t)h′(t)

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Implicit differentiation

The composite function rule may also be used for differentiating functions expressed inan implicit form. For example, let us assume that we have

y3 = x2

To determine the derivative dydx we use implicit differentiation: Treat y as an unknown

function of x and differentiate both sides term by term with respect to x. This gives

ddx

(y3) =ddx

(x2)

Now y3 is a composite function of x with y being the intermediate variable, so thecomposite function rule gives

ddx

(y3) =ddy

(y3)dydx

= 3y2 dydx

Then substituting back

3y2 dydx

= 2x ordydx

=2x3y2

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Outline

1 Limits





6 Optimum Values


PDL, RD, IS (CoE) WS 2016 26/ 40


Higher Derivatives

The derivative df/dx of a function f (x) is itself a function and may be differentiable.

The derivative of a derivative is called the second derivative and is written as

d2fdx2

or f ′′(x) or f (2)(x) or D2f

To obtain it we simply differentiate df/dx again with respect to x

d2fdx2

=ddx

(dydx

)The second derivative may itself be differentiated, yielding third derivatives and so on.

In general the nth derivative is written as

dnfdxn

or f (n)(x) or Dnf

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Example

QuestionFind the second and third derivatives of y = sin 2xSolutionWe first determine the first derivative

dydx

= 2 cos 2x

We differentiate this to get the second derivative

d2ydx2

=ddx

(dydx

)= −4 sin 2x

Differentiating again yields the third derivative

d3ydx3

=ddx

(d2ydx2

)= −8 cos 2x

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Properties of the Second Derivative

The second derivative d2f/dx2 represents the rate of change of df/dx, this gives usinformation on how the slope of the tangent changes with increasing x.

If d2f/dx2 > 0 then df/dx is increasing as x increases, and the tangent rotates inan anti clockwise direction as we move along the horizontal axis resulting in agraph which is concave up.

If d2f/dx2 < 0 then df/dx is decreasing as x increases and the tangent rotates in aclockwise direction as move along the horizontal axis resulting in a graph which isconcave down.

Concave up

x

y

y=f(x)

Concave down

x

y

y=f(x)

d2f/dx2 > 0 d2f/dx2 < 0

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Outline

1 Limits





6 Optimum Values


PDL, RD, IS (CoE) WS 2016 30/ 40


In many situations in engineering we are interested to find optimum values eg themaximum or minimum of a function. This generally occurs when the tangent ishorizontal and

dfdx

= f ′(x) = 0

x

Local Minimum

Local Maximumy=f(x)

Absolute Maximum

Absolute Minimum

y

However, the test f ′(x) = 0 usually leads to points that are usually only local maximaand local minima of the function. In seeking extremal values we need to also checkthe end points (if any) of the domain of the function.

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Stationary Points

We call the points which have f ′(x) = 0 stationary or critical points. As well aspossibly being local maxima and local minima these points might also be points ofinflection.

A point of inflection is where the graph crosses its own tangent, which might be behorizontal. In the case where the tangent is horizontal we call it a stationary point ofinflection.

y=f(x)

y

x

Point of inflection

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Other Exceptions

Note that there are also functions which have optimum values at points other thanstationary points.

This can happen when the function is not differentiable at a point. A simple example ofthis is the function f (x) = x2/3, which has a minimum at x = 0 but as it does not have aunique tangent at this point is not differentiable at this point.

0

0.5

1

1.5

2

2.5

3

-4 -3 -2 -1 0 1 2 3 4

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Method 1 for Determining the Nature of Stationary Points

To find local maxima and local minima

Determine the location of the stationary points where f ′(x) = 0.

If the value of f ′(x) changes from positive to negative as we pass from left to rightthrough a stationary point then the latter corresponds to a local maximum.

If the value of f ′(x) changes from negative to positive as we pass from left to rightthrough a stationary point then the latter corresponds to a local minimum.

To find points of inflection

If f ′(x) does not change sign as we pass through a stationary point then the lattercorresponds to a point of inflection. In particular we usually call this a stationarypoint of inflection as in this case f ′(x) = 0 at the point inflection. However pointsof inflection can occur at other locations too..

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Method 2 for Determining the Nature of Stationary Points

To find local maxima and minima

Find the stationary points x = a where f ′(x) = 0.

If f ′′(x) < 0 the function has a local maximum at x = a.

If f ′′(x) > 0 the function has a local minimum at x = a.

To find points of inflection

Find the points x = b where f ′′(b) = 0

Check whether f ′′(x) changes sign as x passes through x = b. If this occurs thenx = b is a point of inflection.

If in addition f ′(b) = 0 then x = b is a stationary point of inflection.

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Example

QuestionUsing method 2 investigate the nature of the stationary points of the function

f (x) = 4x3 − 21x2 + 18x + 6

Solution

-40

-20

0

20

40

60

80

-1 0 1 2 3 4 5

f(x)

x

The derivative of this function is

f ′(x) = 12x2 − 42x + 18

Stationary points are where f ′(x) = 0 and solving the quadratic gives (0.5, 10.25) and(3,−21). The second derivative of the function is

f ′′(x) = 24x− 42

At the stationary point (0.5, 10.25), f ′′(0.5) = −30 < 0 and so this corresponds to alocal maximum. At the stationary point (3,−21), f ′′(3) = 30 > 0 and so thiscorresponds to a local minimum. Note that f ′′(x) = 0 at x = 1.75 and that f ′′(x) < 0 forx < 1.75 and f ′′(x) > 0 for x > 1.75. Thus the point (1.75,−5.375) is a point ofinflection, but not a stationary point of inflection.

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Outline

1 Limits





6 Optimum Values


PDL, RD, IS (CoE) WS 2016 37/ 40


L’Hopital’s Rule

L’Hopital’s rule allows us to compute limits of the form

limx→a

f (x)

g(x)

in the case where f (a) = g(a) = 0. In this case

limx→a

f (x)

g(x)=

f ′(a)

g′(a)

provided that g′(a) 6= 0.

It may be that f ′(a)g′(a) is also indeterminate. In such cases we must repeat the process of

applying L’Hopital’s rule each time we have 00 at x = a. We must stop through if one or

more of the derivatives is non-zero.

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Applications of L’Hopital’s Rule

ExampleUsing L’Hopital’s Rule determine the limit

limx→0

sin x− xx3

SolutionSince sin x−x

x3 is indeterminate at x = 0 we apply L’Hopital’s Rule

limx→0

sin x− xx3

= limx→0

cos x− 13x2

again 0/0 at x = 0

= limx→0

− sin x6x

again 0/0 at x = 0

= limx→0

− cos x6

= −16

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Applications of L’Hopital’s Rule

ExampleUsing L’Hopital’s Rule determine the limit

limx→0

1− cos xx + x2

SolutionSince 1−cos x

x+x2 is indeterminate at x = 0 we apply L’Hopital’s Rule

limx→0

1− cos xx + x2

= limx→0

sin x1 + 2x

= 0

Note in this case the limit is zero as sin x1+2x has the form of 0/1 at x = 0. If we mistakingly

proceeded we would get

limx→0

1− cos xx + x2

= limx→0

sin x1 + 2x

= limx→0

cos x2

=12

which is incorrect.

PDL, RD, IS (CoE) WS 2016 40/ 40

engineering analysis 1 : differentiation · we use implicit differentiation: treat y as an unknown...

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