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ACCESS TO ENGINEERING - MATHEMATICS 2
ADEDEX
SEMESTER 2 2014/
DR. ANTHONY BROWN
- Differential Calculus
3.1. Introduction to Differentiation.
For most of this chapter we will concentrate on the mechanics of how to differentiate functions but before that we will quickly revise what it means to find the derivative of a function and what that derivative represents.
We will start with the formal definition.
Definition 3.1.1 (Derivative). Let f : (a, b) → R, then the derivative of f at x ∈ (a, b) is defined to be
f ′(x) = lim h→ 0
f (x + h) − f (x) h
if this limit exists.
Before we go any further let us examine what this definition means. If we look at Figure 1, we will see that the slope of the line from the point (x, f (x)) to the point
(x + h, f (x + h)) is just
f (x + h) − f (x) h
When we differentiate a function at a point x, what we are really doing is to make
h smaller and smaller in
f (x + h) − f (x) h
and see what happens to it. What this
means graphically is shown in Figure 2. Hopefully this figure will convince you that the derivative of f at x is the slope of the tangent line to the function f at the point x.
Remark 3.1.2. Often the derivative of a function f will be denoted by
dy dx
df dx
d dx
(f ) or even fx rather than f ′(x). Also note that sometimes completely different
letters may be use, so you may see things like g′(x) or
dg dx
or indeed
dx dy
, where the
roles of x and y have been reversed. All these different notations mean exactly the same thing. They only exist since calculus was developed by different mathemati- cians and the various notations have persisted.
Figure 1. Slope of a line connecting (x, f (x)) to (x + h, f (x + h)).
Figure 2. Geometric meaning of the derivative.
While all this may seem like a lot of effort to go to just to find the gradient of a tangent to a curve, it is extremely important since it arises in numerous different areas. Whenever you want to find the rate of change of something then calculus will come in handy. For example, if you have a function representing the position of an object, then the derivative will represent the velocity of the object. Similarly if you have a function representing the velocity of an object then the derivative of this function will represent the acceleration of the object.
f (x) f ′(x) Comments 0 0 Note the derivative of 0 is 0 2 0 − 4 0 −π 0 −π is just a number e 0 e is just a number cos(1) 0 cos(1) is just a number x 1 Since x = x^1 , n = 1 giving 1x^0 = 1 x^3 3 x^2 Here we take n = 3 x−^4 − 4 x−^5 = −
x^5
Here we take n = − 4 xπ^ πxπ−^1 π is just a number x−e^ −ex−e−^1 = −
e xe+^
e is just a number ex^ ex^ Here we take a = 1 e^5 x^5 e^5 x^ Here we take a = 5 e−^7 x^ − 7 e−^7 x^ Here we take a = − 7 eex^ e · eex^ = eex+1^ Here we take a = e ln(x)
x
Here we must have x > 0
ln(5x)
x
Here we must have x > 0
ln(− 5 x)
x
Here we must have x < 0 sin(x) cos(x) Here we take a = 1 sin(3x) 3 cos(3x) Here we take a = 3 sin(− 2 x) −2 cos(− 2 x) Here we take a = − 2 sin(−πx) −π cos(−πx) Here we take a = −π cos(x) − sin(x) Here we take a = 1 cos(4x) −4 sin(4x) Here we take a = 4 cos(− 5 x) 5 sin(− 5 x) Note −(−5) = + cos(πx) −π sin(πx) Here we take a = π Table 2. Some examples of derivatives
Here are a couple of examples of the use of the Sum Rule.
Example 3.3.2. (1) Find the derivative of f (x) = x^2 + sin(2x).
f ′(x) =
d dx
(x^2 ) +
d dx
(sin(2x)) = 2x + 2 cos(2x).
(2) Find the derivative of f (x) = ln(2x) + e−^3 x. Provided x > 0 (so that the derivative of the first term exists),
f ′(x) =
d dx
(ln(2x)) +
d dx
(e−^3 x) =
x
− 3 e−^3 x.
The second rule that will enable us to differentiate a larger range of functions is the Multiple Rule.
Theorem 3.3.3 (The Multiple Rule for Differentiation). Let f : (a, b) → R and c ∈ R, then the derivative of cf at x ∈ (a, b) is given by
(cf )′(x) = cf ′(x),
provided the derivative of f exists.
All this says is that if we want to differentiate a constant multiple of a function, then all we have to do is first differentiate the function and then multiply by the constant.
Here are a couple of examples of the Multiple Rule.
Example 3.3.4. (1) Find the derivative of f (x) = 5x^3.
f ′(x) = 5 ×
d dx
(x^3 ) = 5 × 3 x^2 = 15x^2.
(2) Find the derivative of f (x) = −3 cos(2x).
f ′(x) = − 3 ×
d dx
(cos(2x)) = − 3 × (−2 sin(2x)) = 6 sin(2x).
Warning 3.3.5. The Multiple Rule can only be used to differentiate a product of a number and a function. If we want to differentiate the product of two functions, then we have to use the Product Rule which is introduced in Section 3.
Of course we are free to use both the Sum and Multiple Rules to differentiate a function and the following are a couple of examples of this.
Example 3.3.6. (1) Find the derivative of f (x) = 5x^2 − 4 x + 3.
f ′(x) =
d dx
(5x^2 ) +
d dx
(− 4 x) +
d dx
(3) (using the Sum Rule)
d dx
(x^2 ) − 4
d dx
(x) +
d dx
(3) (using the Multiple Rule)
= 5(2x) − 4(1) + 0 = 10x − 4. (2) Find the derivative of f (x) = −e−^2 x^ − 2 cos(− 3 x).
f ′(x) =
d dx
(−e−^2 x) +
d dx
(−2 cos(− 3 x)) (using the Sum Rule)
d dx
(e−^2 x) + (−2)
d dx
(cos(− 3 x)) (using the Multiple Rule)
= −(− 2 e−^2 x) − 2(−3(− sin(− 3 x))) = 2e−^2 x^ − 6 sin(− 3 x).
(2) Find the derivative of f (x) = 2 sin(2x) cos(−x) − 3 e−^2 x^ ln(4x). Provided x > 0 (so that the derivative of the ln(4x) exists),
f ′(x) =
d dx
(2 sin(2x) cos(−x)) +
d dx
(− 3 e−^2 x^ ln(4x)) (using the Sum Rule)
d dx
(sin(2x) cos(−x)) − 3
d dx
(e−^2 x^ ln(4x)) (using the Multiple Rule)
d dx
(sin(2x)) cos(−x) + sin(2x)
d dx
(cos(−x))
d dx
(e−^2 x) ln(4x) + e−^2 x^
d dx
(ln(4x))
(using the Product Rule)
=2 (2 cos(2x) cos(−x) + sin(2x)(−(− sin(−x))))
− 3
− 2 e−^2 x^ ln(4x) + e−^2 x^
x
=4 cos(2x) cos(−x) + 2 sin(2x) sin(−x) + 6e−^2 x^ ln(4x) − 3 e−^2 x^
x
3.5. The Quotient Rule.
The next rule that we will need is the Quotient Rule, which will enable us to differ- entiate a function which may be regarded as a quotient of two simpler functions.
Theorem 3.5.1 (The Quotient Rule for Differentiation). Let f : (a, b) → R and
g : (a, b) → R, then the derivative of
f g
at x ∈ (a, b) is given by
f g
(x) =
f ′(x)g(x) − f (x)g′(x) g^2 (x)
provided these derivatives exist and provided that g(x) 6 = 0.
I don’t think it adds anything to express this theorem in words but a couple of examples will be helpful.
Example 3.5.2. (1) Find the derivative of f (x) = tan(x) =
sin(x) cos(x)
Provided cos(x) 6 = 0,
f ′(x) =
d dx
(sin(x)) cos(x) − sin(x)
d dx
(cos(x)) cos^2 (x)
=
cos(x) cos(x) − sin(x)(− sin(x)) cos^2 (x)
=
cos^2 (x) + sin^2 (x) cos^2 (x)
=
cos^2 (x) = sec^2 (x).
(2) Find the derivative of f (x) =
ex x^2
f ′(x) =
d dx
(ex)x^2 − ex^
d dx
(x^2 ) (x^2 )^2
=
exx^2 − ex(2x) x^4
=
exx^2 − 2 xex x^4
Of course, the Quotient Rule can also be combined with the Sum, Multiple or Product Rules as necessary.
3.6. The Chain Rule.
The last rule that we will need is the Chain Rule (also called the Composition Rule). This enables us to differentiate functions of the form f (x) = sin(x^3 ) where there is an ‘outer’ function (in this case sin(x)) which is a function of an ‘inner’ function (in this case x^3 ).
Theorem 3.6.1 (The Chain Rule for Differentiation). Let u be a function of x and y be a function of u, then the derivative of y = f (x) is given by
dy dx
dy du
du dx
provided these derivatives exist.
Remark 3.6.2. This theorem can also be written in notation similar to the previous theorems in this section (as (f ◦ g)′(x) = (f ′^ ◦ g)(x) · g′(x)). However I don’t recommend this latter equation when actually differentiating, it is much easier to use the equation in the theorem.
Thus the critical points of f are x = 1 and x = 2. These are shown in Figure 3.
Figure 3. Critical points of the function f (x) = 2x^3 − 9 x^2 + 12x + 5.
(2) Find the critical points of f (x) = cos(x). Here f ′(x) = − sin(x), so the critical points of f are the points where − sin(x) = 0, that is where sin(x) = 0. Thus the critical points are x = kπ, where k is any integer. Some of these are shown in Figure 4.
Figure 4. Some of the critical points of the function f (x) = cos(x).
Remark 3.7.3. In Example 3.7.2.1, f had two critical points while in Example 3.7.2.2, f had an infinite number of critical points. In general a function can have any number of critical points (including 0).
3.8. Finding Maxima and Minima.
Critical points are important since functions often attain their maximum or mini- mum at them. However they are not the only places where functions attain maxima and minima and so, before we go on to describe how to classify critical points, we will describe two different sorts of maximum and minimum and give another method of finding them. We will first define exactly what we mean by maxima and minima.
Definition 3.8.1 (Global maximum). Given a set S and a function f : S → R, then we say f attains a global maximum at a ∈ S if f (x) 6 f (a) for all x ∈ S.
Definition 3.8.2 (Global minimum). Given a set S and a function f : S → R, then we say f attains a global minimum at a ∈ S if f (a) 6 f (x) for all x ∈ S.
Definition 3.8.3 (Local maximum). Given a set S and a function f : S → R, then we say f attains a local maximum at a ∈ S if there exists some number b such that (a − b, a + b) ⊆ S and such that f (x) 6 f (a) for all x ∈ (a − b, a + b) ∩ S.
Definition 3.8.4 (Local minimum). Given a set S and a function f : S → R, then we say f attains a local minimum at a ∈ S if there exists some number b such that (a − b, a + b) ⊆ S and such that f (a) 6 f (x) for all x ∈ (a − b, a + b) ∩ S.
Remark 3.8.5. Note the use of the 6 symbol in the above definitions. If we replace 6 with <, then we say that the maxima or minima are strict.
As usual, a diagram will make it more obvious what all these definitions mean. All the global and local maxima and minima of the function
f : [0, 3] → R x 7 → 2 x^3 − 9 x^2 + 12x + 5
are shown in Figure 5.
Figure 5. Global and local maxima and minima of the function f (x) = 2x^3 − 9 x^2 + 12x + 5 with domain [0, 3].
points and at the endpoints of the domain. To see how this works in practice let us look at two examples involving the function with rule f (x) = 2x^3 − 9 x^2 + 12x + 5 but with two different domains.
Example 3.8.6. (1) Find the global maxima and minima of the function
f : [0, 3] → R x 7 → 2 x^3 − 9 x^2 + 12x + 5. We already know from Example 3.7.2.1 that the critical points of f are x = 1 and x = 2 (and these are in the domain). The endpoints of the domain are x = 0 and x = 3, so to find the global maxima and minima, we evaluate f at the above four points. f (0) = 5, f (1) = 10, f (2) = 9 and f (3) = 14. The smallest of these numbers is 5, so the global minimum of f occurs at x = 0. The largest of these numbers is 14, so the global maximum of f occurs at x = 3. (2) Find the global maxima and minima of the function f : [0. 8 , 2 .2] → R x 7 → 2 x^3 − 9 x^2 + 12x + 5. The only difference here is that the endpoints of the domain are x = 0. 8 and x = 2.2, so we use these two points instead of x = 0 and x = 3. f (0.8) = 9.864, f (1) = 10, f (2) = 9 and f (2.2) = 9.136. The smallest of these numbers is 9, so the global minimum of f occurs at x = 2. The largest of these numbers is 10, so the global maximum of f occurs at x = 1.
3.9. Classifying Critical Points.
It may be that we are only interested in looking at the behaviour of a function near a critical point and this section gives one method of determining this in many cases. Since the derivative of a function is also a function, we can differentiate it again.
Definition 3.9.1 (Second Derivative). The second derivative of a function f is defined to be the derivative of f ′(x).
Remark 3.9.2. The second derivative of a function f may be denoted by f ′′(x) or
f (2)(x) or
d^2 f dx^2
or
d^2 y dx^2
or fxx
Theorem 3.9.3 (The Second Derivative Test). Let S be a set and let f : S → R have a critical point at a ∈ S. Then there is a local maximum at a if f ′′(a) < 0 and there is a local minimum at a if f ′′(a) > 0.
Warning 3.9.4. Note that the test does not mention the case when f ′′(a) = 0. If it does happen that f ′′(a) = 0 then the test does not tell us anything at all. In this case we can NOT conclude that f does not have a local maximum or local minimum at the point, it just means that the test doesn’t work! For example, if f (x) = x^4 , then f ′′(0) = 0 but f has a local minimum at x = 0. Similarly if f (x) = −x^4 , then f ′′(0) = 0 but f has a local maximum at x = 0.
Remark 3.9.5. In fact the points where f ′′(x) = 0 have a special name, they are called points of inflection. They may or not be critical points. For example f (x) = sin(x) has a point of inflection at x = 0 but this is not a critical point (f ′(0) = cos(0) = 1 6 = 0 while f ′′(0) = − sin(0) = 0).
Let us now give an example to see how the Second Derivative Test works with our function f (x) = 2x^3 − 9 x^2 + 12x + 5 (where this time we can take the domain to be R).
Example 3.9.6. Classify the critical points of f (x) = 2x^3 − 9 x^2 + 12x + 5. We already know from Example 3.7.2.1 that f ′(x) = 6x^2 − 18 x + 12 and that f has critical points at x = 1 and x = 2. Now to find the second derivative we differentiate f ′. We have f ′′(x) = 12x − 18, so that f ′′(1) = −6 and f ′′(2) = 6. Since f ′′(1) < 0, f has a local maximum at x = 1 and since f ′′(2) > 0, f has a local minimum at x = 2.
3.10. The Newton-Raphson Method.
In the final section of this chapter, we will use calculus to find approximate solutions to equations. In fact it is numerical methods like this that computer algebra packages (such as Maple and Mathematica) and indeed some programmable calculators use to solve equations.
The method enables us to find an approximate solution to the equation f (x) = 0 (and note we can always put an equation into this form by bringing all the terms over to one side of the equation). If we have an sufficiently good estimate (I won’t go into what this means in this course) of the solution, say, xn, then a better estimate of the solution is given by
(1) xn+1 = xn −
f (xn) f ′(xn)
So what we do is to make a first guess at the solution, say x 0 , and then successively get better ones x 1 , x 2 , etc. Figure 8 shows the sort of thing that happens (if we are lucky).
Warning 3.10.1. Above I said ‘if we are lucky’, since other things can happen. The iterations could tend to a solution we don’t want or they could tend to infinity. In this course I will just ask you to perform so many iterations but please do be aware that things don’t always work out nicely in practice and there is a huge area of Mathematics that studies iterative methods like this - such as the area that studies fractals like the Mandelbrot set.
