MA10192: Mathematics I semester 1

Transcription

1 MA09: Mathematics I 0-0 semester

2 Contents Functions and equations 3. Polynomials Exponentials and logarithms Trigonometric functions Definitions Solving simple trigonometric equations Inverse trigonometric functions Polar coordinates Harmonic form Solving more complicated trigonometric equations The bisection method Differentiation 8. Definition Notation Derivatives of standard functions Rules for differentiation Product rule Quotient rule Chain rule Derivatives of inverse functions Parametric differentiation Implicit differentiation Maxima and minima Asymptotes L Hopital s rule Newton s method Taylor polynomials Numerical differentiation Integration Antiderivatives Estimating area Definition (optional)

3 3.4 The fundamental theorem of calculus Rules for integration The substitution rule Integration by parts Anti-derivatives of inverse functions Anti-derivatives of rational functions by partial fractions Anti-derivatives of trigonometric rational functions Trigonometric substitution Hyperbolic substitution Area between curves Numerical integration Improper integrals Ordinary differential equations Separable equations Linear equations Integrating factors Variation of parameters Method of undermined coefficients Numerical methods Functions of several variables Partial derivatives and extrema Lagrange multipliers and boundary extrema The chain rule Exact differential equations

4 Chapter Functions and equations. Polynomials The simplest kind of functions are linear ones. The formula is y = ax + b, and geometrically this is a straight line (with slope a and y-intercept b). Finding the point of intersection of the straight lines y = ax + b and y = cx + d leads to the linear equation ax + b = cx + d. We rewrite this as (a c)x = d b and note that if a c this can be solved as x = d b a c, whereas if a = c and d b there is no solution (geometrically: the lines are parallel and not equal) and if a = c and d = b every x is a solution (geometrically: the lines are equal). We find the y-coordinate of the point of intersection by substituting the found x in either formula: y = a d b a c + b. Given a line L and a point P we can define a parabola as the set of points (x, y) such that the distance of (x, y) to P is the same as the distance of (x, y) to L. If we for example take P = (0, ), then the distance of (x, y) to (0, ) Actually, this formula gives what is more properly called an affine function, the function is linear only if b = 0. In the one variable case this misuse of terminology is very widespread. But do remember the distinction when studying linear functions of several variables in Mathematics II. 3

5 is (by Pythagoras Theorem) x + (y ). And if we take L to be the line y =, then the distance of (x, y) to this line is y + (i.e. the absolute value of y + ). So for the parabola defined by P = (0, ) and the line L given by y = we obtain the formula x + (y ) = y +. Squaring both sides and simplifying gives the formula y = 4 x. We will not go further into the geometry of parabolas in this course. Suffice it to say that the formula y = ax + bx + c always defines a parabola and that any parabola may be written in this way by choosing the coordinate axis in the right way (the crucial thing for the latter is to take the x-axis parallel to the given line). Finding the point of intersection of parabolas or a parabola and a straight line gives rise to a quadratic equation ax + bx + c = 0. We will tacitly assume that a 0 since otherwise this really is a linear equation, which we already saw how to solve. We first look at the special case where b = 0, a =. We rename c = z. We then get the equation x = z. Geometrically it is clear that if z > 0, then there are two solutions, one positive and one negative: it is also clear the the negative solution is minus the positive solution. The positive one is called z. Note that we haven t really solved this equation, we simply noted that there is a solution and have given that solution a name. For certain values of z we know what z is; for example if z = 4, then its square root is. But for example cannot be simplified; it is not a whole number or a quotient of whole numbers. We now reduce the general quadratic equation to the special quadratic equation x = z. We have ( [ ax + bx + c = a x + b ] ) + b + 4ac a 4a so that the solutions of ax + bx + c = 0 are x = b ± b 4ac. a This is the well-known quadratic formula. Note that this expresses the solution of the general quadratic equation ax + bx + c = 0 in terms of those of a special quadratic equation, namely x = z (which has square roots as its solutions). From the graph of the function y = x 3 it is obvious that the equation x 3 = z has exactly one solution for any real number z. Similarly as before with quadratic equations we may ask the question whether the general cubic equation ax 3 + bx + cx + d = 0 can be solved in terms of cube roots (solutions 4

6 of x 3 = z) and maybe square roots as well. This can indeed be done, but the formulas are not very nice. Quartic equations (ax 4 +bx 3 +cx +dx+e = 0) can be solved as well using only roots. Surprisingly, it stops there: quintic equations cannot be solved in terms of roots (a specific example is x 5 x + = 0). This is the content of the Abel-Ruffini theorem. We return to the equation x = z. As said, for z > 0 this equation has two solutions. For z = 0 obviously only zero is a solution. For z < 0 there are again two solutions, but this are not real numbers but complex numbers. You might know them from A-level, they will appear in MA093: Mathematics II.. Exponentials and logarithms You are undoubtedly familiar with the exponential function a x with a > 0. There are various ways of formally defining these functions. The most elementary and most useful at this point is the following. An exponential function is a continuous function f such that f(0) = and such that for all x and y the following holds f(x + y) = f(x)f(y), (.) i.e. exponential functions are those that translate addition into multiplication. The continuity assumption is made to rule out pathological examples and we will not worry too much about it (intuitively a continuous function is a function whose graph you can draw without taking your pen from the paper ). It can be shown that an exponential function is completely characterized by its value at some point different from zero. This point is usually taken to be one and the exponential function is then written as a x with a the value at one. The formula (.) then becomes a x+y = a x a y, (.) which should be familiar. We can t go into the details of this (existence and uniqueness of exponential functions), but note that a x certainly makes sense for x a nonnegative integer and since a z/ a z/ = a z by (.) with x = y = z/ we must have a / = a, and we already encountered square roots. Similarly, for all rational numbers x (i.e. fractions, i.e. quotients of integers a x can be defined in terms of roots. It can then be shown that a x is continuous and it can subsequently be extended by continuity to all irrational numbers (such as and π) as well. We will really only be interested in solving equations involving exponential functions (and later in differentiating them and integrating them). So don t worry if this was a bit above your head, the formula (.) is something that you should definitely keep in mind though. Another one to remember is (a x ) y = a xy, (.3) The precise defintion is as follows: f is continuous at the point p if for all ε > 0 there exists a δ > 0 such that if p δ < x < p + δ, then f(p) ε < f(x) < f(p) + ε. A function is continuous if it is continuous at all points where it is defined. 5

7 i.e. powers are turned into products. Note that if y is a positive integer, then this easily follows from (.) (it follows in the general case as well, but not easily). Assuming that a > 0 and a, the equation a y = x can be solved for y if x > 0 and such a solution y is unique (remember the graph of the exponential function). This solution y is called the a-logarithm of x: log a x. So by definition of the logarithm we have for all x > 0 We also have (now for all x): a log a x = x. (.4) log a a x = x. (.5) The equations (.4) and (.5) mean that the exponential and logarithm are inverses of each other. We can use the identities (.) and (.3) to obtain the following identities for logarithms: log a (xy) = log a x + log a y, log a (x y ) = y log a x, (.6) so the logarithm turns products into sums and powers into products. If you ve seen one exponential function then you ve seen them all because of the equation b x = c x log c b, which we obtained by using b = c log c b from (.4), then taking both sides of that equation to the power x and using (.3). So if we would only know how to compute c-logs and c-exponentials, then we can still compute b-exponentials for any b. You ve probably used the corresponding formula for logarithms log b x = log c x log c b when using your calculator (which probably only computes 0-logarithms and e- logarithms). So we really only need to consider the exponential and logarithmic function for one fixed base. For several reasons the base e (.788) is the best base. The reason you will probably find most convincing is that the exponential function a x has a derivative equal to a constant times itself and that only for a = e this constant equals one (so that e x is the only exponential function that is its own derivative). We will come back to this when we discuss differentiation. The natural logarithm log e at A-level is often denoted by ln and I will use this notation as well. But be warned that many people write log when they mean the natural logarithm (though your calculator probably uses the symbol log for log 0 ). Now, let us solve equations. 6

8 Problem. The following expression is an integer, which integer? 9 5/ 8 3/4 7 /3. Solution. We use the exponential rules (.) and (.3) in the following computation: 9 5/ 8 3/4 7 /3 = (3 ) 5/ (3 4 ) 3/4 (3 3 ) /3 = = 38 3 = 36 = 79. Problem. Solve for x: 5 3x =. Solution. There are at least three ways of solving this problem. They give solutions that look different, but are actually the same. The first method is the following.take the log 5 on both sides to obtain, using (.5), 3x = log 5 and divide by 3 to isolate x: x = log 5. 3 The second method is as follows. Note that 5 3x = ( 5 3) x = 5 x by (.3) so that the equation is equivalent to 5 x =, which by the definition of the logarithm has solution x = log 5. The third method, which I prefer since it gives the answer in terms of natural logarithms goes as follows. Take the natural logarithm of both sides of the equation to give ln 5 3x = ln. Now use the logarithmic rule (.6) to rewrite this as 3x ln 5 = ln. From this we solve x as x = ln 3 ln 5. Using any of these three methods of solving the equation is fine. Problem 3. Solve for x: ln 5x =. Solution. Take e. on both sides to obtain, using the definition of logarithm (.4), and divide by 5 to isolate x: 5x = e x = e 5. 7

9 Problem 4. Solve for x: ln e 3x = 7. Solution. The left-hand side is 3x (using (.5)), so x = 7/3. Problem 5. Solve for x: ln (x + ) + ln (x + ) = 5. Solution. Take e. on both sides to obtain e ln (x+)+ln (x+) = e 5. Now use the exponential rule (.) to obtain e ln (x+) e ln (x+) = e 5. Use the relation between exponential and logarithm (.4) to obtain (x + )(x + ) = e 5. Write this second order equation in standard form Use the quadratic formula and simplify a bit x + 3x + e 5 = 0. x = 3 ± 9 4( e 5 ), x = 3 ± + 4e 5. If we choose the minus sign, then x is not in the domain of the function ln (x + ) + ln (x + ) that we started with (this domain consists of all numbers larger than -). So only the plus sign gives a valid solution: x = e 5. Problem 6. Solve for x: Solution. Rewrite as Substitute u = x : Rewrite 4 x + x = 0. ( x ) + x = 0. u + u = 0. (u + )(u ) = 0. So u = or u =. So we have to solve x = and x =. The first has no solution, the second the unique solution x = 0. 8

10 Problem 7. Suppose the function f(x) = ca x satisfies f() = 3, f(4) = 5. What are c and a? Solution. We get the equations We eliminate c by taking the quotient: ca = 3, ca 4 = 5. ca 4 ca = a3 = 5 3. So a = ( 5 3) /3. Substitute back to obtain so c ( ) /3 5 = 3, 3 c = 3 ( ) / Problem 8. Prove, using the exponential laws that for all x, y > 0 ln x y = ln x ln y. Solution. Take e. on both sides gives the equivalent First, we have so that e ln y = y. So we indeed have x y = eln x ln y = e ln x e ln y. = e 0 = e ln y e ln y = y e ln y, e ln x e ln y = x y..3 Trigonometric functions.3. Definitions We now define the trigonometric functions sine and cosine in a maybe at first sight slightly odd way. Consider the unit circle (i.e. all pairs (x, y) such that x + y = ). Let t be some number greater than or equal to zero. Starting from the point (, 0) walk along the unit circle counter clock-wise for length t. 9

11 Define cos t as the x-coordinate of the point that you arrive at and sin t as the y-coordinate of the point that you arrive at. If t < 0 do a similar thing but now walk clockwise for length t. The so defined functions obviously have certain properties: Periodicity sin(t + π) = sin t, cos(t + π) = cos t, Sine is odd sin( t) = sin t, Cosine is even cos( t) = cos t, Pythagoras cos t + sin t =. There are also the following, not so obvious but important, addition formulas: sin(s + t) = sin s cos t + cos s sin t, cos(s + t) = cos s cos t sin s sin t. Note that the addition formulas imply the following properties (which are geometrically reasonably obvious): sin(π t) = sin t, sin( π + t) = cos t, cos( 3π + t) = sin t. We define the third trigonometric function, the tangent, as tan t = sin t cos t. It is possible to define the tangent geometrically, but the above definition will do for our purposes. The other trigonometric functions secant, cosecant and cotangent can also be defined geometrically, but I will ignore these functions in this course..3. Solving simple trigonometric equations It seems that at A-level trigonometric equations only have to be solved on a given finite interval. I will virtually always ask you to find all solutions. For example, if I ask you to solve sin x = 0 for x, then I expect the answer x = kπ with k an integer. I might occasionally ask you to solve sin x = 0 for 3π x π and then the correct answer would be x = 3π, π, π, 0, π, π. This can be deduced from the general case by noting that x = kπ with k an integer is in the desired interval if and only if k = 3,,, 0,,. Problem 9. Solve cos x = for x. Solution. One solution is x = π 3. Since the cosine is even, another solution is x = π 3. Geometrically it is obvious that these are the only two solutions that satisfy π x π (one full period). Since the cosine is π-periodic all solutions are given by x = π 3 + kπ, x = π 3 + nπ, where k and n are arbitrary integers. 0

12 Problem 0. Solve cos x = for x. Solution. We substitute u = x to reduce this problem to the previous one: cos u =. From problem 9 we know that u = π 3 + kπ, u = π 3 + nπ. Now substitute back to obtain x = π 3 + kπ, x = π 3 + nπ. Solving for x gives x = π 6 + kπ, x = π 6 + nπ with k and n are arbitrary integers. Note that we reduced to problem involving cos x to a standard problem involving cos u with u = x and after finding u solved for x. This is usually the best way to proceed: it is very easy to make mistakes with trigonometric functions of different periods than the standard π. Problem. Solve sin x = for x. Solution. One solution is x = π 6. Using the identity sin(π x) = sin x we see that π π 6 = 5π 6 is another solution. Geometrically it is obvious that these are the only two solutions that satisfy 0 x π (one full period). Since the sine is π-periodic all solutions are given by x = π 6 + kπ, x = 5π 6 + nπ, where k and n are arbitrary integers. A often made mistake in problems such as problem (involving the sine) is that a wrong second solution is found. Please look at the geometry of the problem to avoid this..3.3 Inverse trigonometric functions The unique solution of cos x = in the interval 0 x π can be expressed in terms of known numbers (namely π 3 ). However, for the unique solution of cos x = 3 no such simple expression exists. Since the unique solution of cos x = 3 in the interval 0 x π is a bit long there is a shorthand for this. For all y with x, the equation cos x = y has a unique solution x in the interval 0 x π. This solution is called the arccosine of y: arccos y. Often the notation cos y is used, but I will never do this since it is easily confused with cos y, which is a different thing altogether. I will actually later in these notes use the notation cos y for cos y since it is consistent with e.g. cos y meaning (cos y), so be advised! Note that instead of the interval 0 x π we could have picked any other interval of length π such that cos x = y has a unique solution of that interval, the interval [0, π] is an arbitrary but customary choice. Problem. Solve cos x = 3 for x. Solution. By definition x = arccos 3 is a solution. Since the cosine is even, x = arccos 3 is also a solution. These are the only solutions on the interval ( π, π], which is one period. So all solutions are given by x = arccos 3 + kπ, x = arccos 3 + nπ, where k and n are arbitrary integers.

13 Figure.: The graph of the arccos Similarly we define the analogue for the sine. For all x with y, the equation sin x = y has a unique solution x in the interval π x π. This solution is called the arcsine of y: arcsin y. Note that the interval on which the arcsine takes its values is different from that on which the arccosine takes its values Figure.: The graph of the arcsin For applications in integration the inverse of the tangent will turn out to be extremely important. For all y, the equation tan x = y has a unique solution x in the interval π x π. This solution is called the arctangent of y: arctan y. Note that whereas the arccosine and arcsine are only defined on the interval [, ], the arctangent is defined on the whole axis..3.4 Polar coordinates Instead of of using cartesian coordinates it is sometimes convenient to use other coordinates. Polar coordinates are the most common example of such other coordinates. Polar coordinates describe a point in the plane by an angle and a

14 Figure.3: The graph of the arctan radius. This is how it works. Consider the circle centered at the origin through the point P that you want the coordinates of. Denote the radius of this circle by r, that is the first polar coordinate (the radius). For the second coordinate measure the length of the circle arc you traverse when going from the point (r, 0) to P and divide this by r. This gives you a number θ in between zero and π, that is the second polar coordinate (the angle) Figure.4: The Cartesian grid Every pair (r, θ) with r > 0 and 0 θ < π (note the strict and the nonstrict inequality) determines a unique point in the plane and conversely each point in the plane except the origin determines such a pair. The origin is special since its angle is not uniquely determined. It is easy to compute the cartesian coordinates of a point given its polar coordinates: (x, y) = (r cos θ, r sin θ). This easily follows from the definition of the sine and cosine. Going the opposite way is slightly more complicated. The radius is easy enough r = x + y by Pythagoras, but the angle is a bit complicated. First remember that z, the absolute value of z is the distance of the point z on the line to zero (so z = z if z 0 and z = z of z < 0). The 3

15 0.5π 0.75π 0.5π π π π.75π.5π Figure.5: The polar grid angle of (x, y) equals arctan y y x if the point lies in the first quadrant, π arctan x if the point lies in the second quadrant, π + arctan y x if the point lies in the third quadrant and π arctan y x if the point lies in the fourth quadrant. It is better to draw a picture each time than to try to memorize these formulas for the angle. in Cartesian coordi- Problem 3. Write the point with radius 4 and angle 3π nates. Solution. The x-coordinate is 4 cos 3π = 0, the y-coordinate is 4 sin 3π = 4. Problem 4. Write (5, 3) in polar coordinates. Solution. The radius is 5 + ( 3) = 34. The find the angle: draw a picture! The angle θ that we are after satisfies θ = π ϕ, where ϕ satisfies tan ϕ = 3 5. So the angle that we are after is π arctan Harmonic form Sometimes we will want to solve equations like 3 sin t + 5 cos t =. For this it helps to write the sum of the sine and cosine of the same period as a phaseshifted cosine, i.e. given A and B to find r and θ such that A sin t + B cos t = r cos(t + θ). 4

16 We use the addition formula for the cosine to expand cos(t + θ). This gives r cos(t + θ) = r cos t cos θ r sin t sin θ. Comparing this to A sin t + B cos t we see that we will need to choose r and θ such that A = r sin θ, B = r cos θ. Note that this is nothing else than writing (B, A) in polar coordinates. So we have A sin t + B cos t = r cos(t + θ) if we choose r the radius and θ the angle of the point (B, A). Please note the minus sign in front of A! Problem 5. Find r and θ such that 3 sin t + 5 cos t = r cos (t + θ) for all t. Solution. We have seen that we should take r, θ as the polar coordinates of (5, 3). These we computed in problem 4: r = 34, θ = π arctan 3 5. Problem 6. Solve 3 sin t + 5 cos t = for t. Solution. We use that 3 sin t + 5 cos t = r cos (t + θ) with r = 34, θ = π arctan 3 5 that we showed in problem 5. So we must solve 34 cos(t + π arctan 3 5 ) =. This gives cos(t + π arctan 3 5 ) = 34. t+π arctan 3 5 = arccos 34 +kπ, t+π arctan 3 5 = arccos 34 +nπ. So t = π+arctan 3 5 +arccos 34 +kπ, t = π+arctan 3 5 arccos 34 +nπ. Here k and n are arbitrary integers..3.6 Solving more complicated trigonometric equations The following problems are slightly more difficult since they involve solving basic trigonometric equations, trigonometric identities and solving polynomial equations. This first one is relatively straightforward since it does not involve using trigonometric identities. Problem 7. Solve cos x + cos x = 0 for x. Solution. Substitute u = cos x. Then the equation becomes u + u = 0. This can be factored as (u + )(u ) = 0. 5

17 So the solutions are So we have to solve u =, u =. cos x =, cos x =. The first of these equations has no solutions. x = nπ with n any integer. The second has as solutions The following problem is somewhat more complicated (and can be done in several ways). Problem 8. Solve cos x + sin x = 0 for x. Solution. The first objective is to write cos x in terms of sin x so to obtain a polynomial equation in sin x. Using the addition formula for the cosine we have cos x = cos x sin x and subsequently using Pythagoras gives cos x = sin x. It follows that cos x = 4 sin x + 4 sin 4 x. So the equation to be solved is equivalent to 4 sin 4 x sin x = 0. Substituting u = sin x this becomes 4u u = 0, which factors as 4u(u ) = 0. So we obtain u = 0 or u =. This gives sin x = 0, sin x = or sin x =. Solving this for x gives x = kπ, x = π 4 + nπ, x = 3π 4 + mπ, x = π 4 + pπ, x = 3π 4 + qπ, where k, n, m, p, q are arbitrary integers..4 The bisection method You may have gotten a different impression at A-level (and in this course up to now...), but it is a rare occasion when equations can actually be solved explicitly. Usually one has to be satisfied with showing that there are solutions (and how many there are) and with only approximating the solutions. The following theorem (the intermediate value theorem) is useful in showing that there are solutions: a continuous function f on the interval [a, b] takes on all values in between f(a) and f(b), i.e. for all y with f(a) y f(b) (or f(b) y f(a) depending on whether f(a) or f(b) is larger) there exists a c with a c b such that f(c) = y. Problem 9. Show that x 3 x + = 0 has a solution with 0 x. Solution. With f(x) = x 3 x + we have f(0) = and f() = 9. By the intermediate value theorem, f takes on all values in between, in particular the value zero. So there indeed exists a solution x with 0 x. We can use the intermediate value theorem to approximate the unique solution of x 3 x + = 0 with 0 x. This is called the bisection method. We evaluate x 3 x + in x = / (the middle of the interval [0, ]). This 6

18 gives 3.875, the only important thing is that this value is negative so the intermediate value theorem tells us that there exists a solution in [0, ]. Next we evaluate the function in 4 (the middle of the new interval), look whether this value is positive or negative and determine from that whether our solution is in the interval [0, 4 ] or in the interval [ 4, ]. In this way we continue halving the size of the interval, so get a better and better approximation of the solution. It s a bit much work to do this by hand, but computers are good in this sort of mindless number crunching. Problem 0. Approximate using the bisection method. Start with an interval whose endpoints are integers and continue until you can be sure that you have the first 3 digits correct. Solution. We use the bisection method to numerically approximate the solution of x =. A reasonable interval to start with seems the interval [, ] since = < and = 4 >. We obtain the following table using the bisection method: x x interval 4 [,].5.5 [,.5] [.5,.5] [.375,.5] [.375,.4375] [.4065,.4375] [.4065,.4875] [.44065,.4875] [.44065, ] So after bisecting 8 times we know for sure that the first three digits of are.4. 7

19 Chapter Differentiation. Definition The idea behind the derivative is that f (x), the derivative of f at the point x, is the slope of the tangent line to the graph of f at x. A tangent line is bit of tricky geometrical object, so we will instead initially consider chords to the graph: i.e. a line that connects two points on the graph. In figure. a function and two chords which both originate from the same point x are drawn Figure.: A graph and two chords When the point to which the chord is drawn comes closer to x, the chord approaches the tangent line to f at x. So it seems reasonable to look at the slope of such chords and look what happens if we take the point to which the chord is drawn closer and closer to x. The line that connects (x, f(x)) and (x + h, f(x + h)) has slope f(x + h) f(x). (.) h We take for f (x) the natural value to complement this expression for h = 0 (note that (.) is meaningless for h = 0). This may sound complicated and 8

20 confusing, but when applied to an example it turns out to not be too difficult. We consider f(x) = x and we want to know the derivative. Now (.) equals (x + h) x h = x + xh + h x h = xh + h h = x + h. Note that the left-hand side is only defined for h 0, but the right-hand side is perfectly well-defined for h = 0. So once we have made these simplifications it is obvious that the natural value to complement this expression for h = 0 is x. So the derivative of x is x (and you probably recall that this is what your A-level teacher said is was). The notation that we use for the natural value to complement (.) for h = 0 is: In general the notation lim h 0 f(x + h) f(x). h lim g(z) = v z w (read as: the limit of g(z) as z tends to w is v) means that v is the natural value for the expression g(z) at w. The derivative may not always exist. Consider the example of the absolute value x (remember that this equals x if x < 0 and x if x 0). See figure. for its graph. It is obvious that (.) equals if h > 0 and if h < 0. This means that there is no altogether obvious value for the limit. So we leave the derivative undefined. Note that this makes geometric sense: there is no tangent line to this curve at x = Figure.: The absolute value function Of course this is still very imprecise. The precise definition is as follows: for all ε > 0 there exists a δ > 0 such if z w δ then g(z) v ε. For the ones who read the previous footnote. Suppose that there does exist a limit v with the desired properties. Then, choosing ε = /, there should exist a δ > 0 such that z δ implies g(z) v /, where g(z) = if z < 0 and g(z) = if z > 0. Now use the triangle inequality g(δ) g( δ) g(δ) v + g( δ) v to conclude that g(δ) g( δ). But g(δ) g( δ) =. This contradiction shows that our assumption that the limit v exists was wrong. So the limit does not exist. 9

21 Problem. Find the tangent line of f(x) = x + 4 in the point (, 5). Solution. We saw that f (x) = x, which means that f () = so that the slope of the desired tangent line is. The tangent line has to pass through the given point. The unique line with these properties is y = (x ) Notation There are various notations in use for derivatives. The derivative of f is denoted by f, f, df dx, f (), Df. The first of these is due to Lagrange, the second one is due to Newton and is mainly used when the independent variable represents time, the third one is due to Leibniz and is very useful in suggesting that the derivative is some kind of quotient, the fourth and fifth ones are mainly useful when dealing with higher order derivatives. The second derivative (i.e. the derivative of the derivative) is denoted by f, f, d f dx, f (), D f. Here you see an advantage of the last three notations over the first two, this is even clearer when dealing with for example the tenth derivative..3 Derivatives of standard functions In the first section we already calculated the derivative of x. Using the binomial theorem it similarly follows that the derivative of x n is nx n for n a nonnegative integer. Actually that formula is true for all numbers n. We will now just show it for n = /, i.e. we will show that the function f(x) = x has derivative f (x) =. In this case we have x f(x + h) f(x) h x + h x =. h The trick is to multiply this fraction by written in somewhat strange form: x + h x h x + h + x x + h + x, carry out the multiplication of the numerators and note that some terms cancel: x + h h + x x + h x x + h h( x + h + h = x) h( x + h + x) =. x + h + x This last expression makes perfect sense for h = 0 and equals. So this is x indeed the derivative. 0

22 We now turn to exponential function. For f(x) = a x we have f(x + h) f(x) h = ax a h a x h = a x ah. h So we see that the derivative of a x equals a constant depending on a but not a on x (namely lim h h 0 h ) times the function ax itself. The number e is the e unique number for which this constant equals one. So lim h h 0 h =. We a will see later on what this constant lim h h 0 h is for general a. We now determine the derivative of the sine function f(x) = sin x. We first determine the derivative in zero. Note that the sine is zero in zero, so that we are interested in sin h lim h 0 h. We use some geometry to show that this limit is. For the moment assume that h satisfies 0 < h π. Consider the point P = (cos h, sin h) and the right-angled triangle with vertices P, (cos h, 0) and (, 0). By Pythagoras we have: distance of P to (cos h, 0) distance of P to (, 0). Since a straight line is the shortest path between two points we also have: distance of P to (, 0) h, the length of the circle arc centered at the origin through P and (, 0). So distance of P to (cos h, 0) h, i.e. sin h h. (.) Now consider the triangle with vertices vertices (, tan h), (, 0) and the origin. Also consider the slice of the unit disc with vertices (cos h, sin h), (, 0) and the origin. Since the disc slice is contained in the triangle we have: area disc slice area triangle. The disc slice is h π of the whole unit disc (which has area π), so its area is h. The area of the triangle is tan h. So we have h tan h. Multiplying this by the positive number cos h h Combining (.) and (.3) gives gives cos h sin h h. (.3) cos h sin h h. (.4) We have only shown this for 0 < h π. However, since the sine function is odd and the cosine is even, the same inequality must hold for π h < 0. See figure.3. As h 0 the left-hand side of (.4) converges to. So we must have sin h h as h 0. So the derivative of the sine in zero is equal to.

23 Figure.3: The functions, sin x x and cos x. Next we obtain the derivative of the cosine in zero. So we want to know what cos h lim h 0 h is. We have by the double angle formula cos h = sin (h/). So Using that lim z 0 sin z z cos h h = sin (h/) = and sin 0 = 0 we have cos h lim = 0. h 0 h sin (h/). h/ So the derivative of the cosine in zero is zero. We now use the addition formulas to obtain the derivative of the sine at an arbitrary point. We have sin (x + h) sin x h = sin x cos h + cos x sin h sin x h = sin x cos h +cos x sin h h h. Letting h 0 and using the calculated values for the derivatives of the sine and the cosine in zero we see that the derivative of sin x is cos x. Similarly, for the cosine we have cos (x + h) cos x h = cos x cos h sin x sin h cos x h so that the derivative of cos x is sin x. Problem. Find the second, third and fourth derivatives of cos x. Solution. = cos x cos h sin x sin h h h, f (x) = sin x, f (x) = cos x, f (3) (x) = sin x, f (4) (x) = cos x.

24 Problem 3. Find the derivative of sin x from first principles. Solution. You probably know what this derivative is from the chain rule that you learned at A-level. Since I asked for first principles you re not allowed to use the chain rule however. We write down the limit that we need to compute sin (x + h) sin x sin (x + h) sin x lim = lim, h 0 h h 0 h use the addition formula split the fraction in two rewrite to get h everywhere sin x cos h + cos x sin h sin x = lim, h 0 h cos h sin h = sin x lim + cos x lim h 0 h h 0 h, cos h sin h = sin x lim + cos x lim h 0 h h 0 h, and substitute t = h to get limits that we know from the results in this section cos t sin t = sin x lim + cos x lim. t 0 t t 0 t These limits we know, the first one is 0 and the second one is. So the derivative of sin x is cos x..4 Rules for differentiation We usually don t use the definition of differentiation to compute derivatives, but use certain rules. The first rule is rather easy and obvious: the derivative of a sum is the sum of the derivatives: (f + g) (x) = f (x) + g (x). Problem 4. Compute the derivative of x + x. Solution. With f(x) = x and g(x) = x we have f (x) = x and g (x) = by the results in the previous section. So the derivative of x + x is x +. 3

25 .4. Product rule The situation for a product is not so straightforward. It turns out that (fg) (x) = f (x)g(x) + f(x)g (x). (.5) Since it is not so obvious that this should be the case, we will prove it. We are interested in f(x + h)g(x + h) f(x)g(x). (.6) h We rewrite this as f(x + h)g(x + h) f(x)g(x + h) + f(x)g(x + h) f(x)g(x) h and split this fraction in two f(x + h) f(x) g(x + h) g(x) g(x + h) + f(x). (.7) h h From the equality of (.6) and (.7), the product rule (.5) follows by letting h 0. Problem 5. Calculate the derivative of x sin x. Solution. Use the product rule to obtain x sin x + x cos x. Problem 6. Calculate the derivative of e x cos x. Solution. Use the product rule to obtain e x sin x + e x cos x..4. Quotient rule The quotient rule is The proof is again based on rewriting. ( ) f (x) = g(x)f (x) g (x)f(x) g g(x). Problem 7. Calculate the derivative of tan x. 4

26 Solution. We have tan x = sin x cos x, and we already know the derivatives of the numerator and denominator. So we can use the quotient rule. d dx cos x cos x + sin x sin x tan x = cos. x Note that this may be rewritten as either cos x or + tan x, depending on how we simplify the fraction. Problem 8. Calculate the derivative of sin x x. Solution. Use the quotient rule to obtain x cos x sin x x. Problem 9. Calculate the derivative of e x x sin x. Solution. First use the product rule to calculate the derivative of the denominator (x sin x) = x cos x + sin x. Use this and the quotient rule to calculate the derivative of the given function. d dx e x x sin x = ex x sin x e x (x cos x + sin x) x sin x..4.3 Chain rule The chain rule is d dx y(u(x)) = y (u(x)) u (x). The chain rule is not so easy to rigorously prove. In the notation of Leibniz the chain rule is dy dx = dy du du dx, and seems obvious, but this just proves that the notation of Leibniz is good for this purpose; it doesn t show that the chain rule is true. The chain rule can be extended in the obvious way to chains of more than two functions. For example in the case of three functions we have dy dx = dy dv du dv du dx. 5

27 Note that I left out where these functions have to be evaluated. This is shorter, but can cause confusion. The appropriate way to write it is dy dy dv (v(u(x))) = (v(u(x))) dx dv du (u(x))du dx (x). Problem 30. Calculate the derivative of e x. Solution. Apply the chain rule with y(u) = e u, u(x) = x; then dy du = eu, du dx = so dy dx = dy du du dx = e x. Problem 3. Calculate the derivative of e cx Solution. Apply the chain rule with y(u) = e u, u(x) = cx; then dy du = eu, du dx = c so dy dx = dy du du dx = cecx. Problem 3. Calculate the derivative of a x. Solution. We have a x = ( e ln a) x = e x ln a ; so with c = ln a in the previous problem we have d dx ax = ln a a x. Problem 33. Calculate the derivative of sin x. Solution. Apply the chain rule with y(u) = sin u, u(x) = x; then dy du du dx = so dy dx = dy du = cos x. du dx = cos u, Problem 34. Calculate the derivative of (x 3 + ) 7. Solution. Apply the chain rule with y(u) = u 7, u(x) = x 3 + ; then dy du = 7u6, du dx = 3x so dy dx = dy du du dx = x (x 3 + ) 6. Problem 35. Calculate the derivative of cos x. 6

28 Solution. Apply the chain rule with y(u) = u, u(x) = cos x; then dy du = u, du dx = sin x so dy dx = dy du = cos x sin x. du dx Problem 36. Calculate the derivative of cos (x ). Solution. Apply the chain rule with y(u) = cos u, u(x) = x ; then dy du = sin u, du dx = x so dy dx = dy du du dx = x sin (x ). Problem 37. Calculate the derivative of e sin (x). Solution. Apply the extended chain rule with y(v) = e v, v(u) = sin u, u(x) = x ; then dy dv = ev, dv du du = cos u, dx = x so dy dx = dy dv du dv du dx (x = ) esin cos (x )x..5 Derivatives of inverse functions If y(x) has an inverse function, i.e. y = y(x) can be solved as x = x(y), and is differentiable, then the inverse function is differentiable and for the derivative of the inverse we have dx dy =. dy dx Note that again the Leibniz notation is very suggestive. We now use this formula to compute the derivatives of the inverse functions of functions of which we previously computed the derivatives. Note that we want dx dy as a function of y. Problem 38. Compute the derivative of x = y /3. Solution. Of course you may use here that the derivative of x n is nx n, but here we will use the formula for the derivative of inverse functions. We have y = x 3, so dy dx = 3x. So dx dy = 3x = 3y /3. Problem 39. Compute the derivative of x = ln y. 7

29 Solution. We have y = e x, so dy dx = ex. So dx dy = e x = y. Problem 40. Compute the derivative of x = arcsin y. Solution. We have y = sin x, so dy dx = cos x. So dx dy = cos x. We want to express this in terms of y. We have cos x = ± sin x = ± y. The question is whether we should have the plus or the minus sign. For this we have to go back to our definition of arcsin. Remember that x = arcsin y is the unique solution of sin x = y with π x π. For these values of x we have cos x 0, so we need the plus sign. So we have cos x = y. That gives dx dy = y. d Similary, dy arccos y =. y Problem 4. Compute the derivative of x = arctan y. Solution. We have y = tan x, so dy dx = tan x +. So dx dy = tan x + = y +..6 Parametric differentiation Sometimes curves are not given as a function y(x), but as pairs (x(t), y(t)), i.e. both coordinates are functions of a parameter t. An example is the unit circle which is given by x(t) = cos t, y(t) = sin t; but cannot be written as the graph of a function y(x). The following formula gives the slope dy dx in terms of the derivatives ẋ and ẏ with respect to the parameter t: dy dx = ẏ ẋ. Writing ẏ and ẋ in Leibniz notation and presuming that one may operator with these expressions as if they are fractions gives an indication (but no proof) of the fact that this formula is indeed valid. 8

30 Problem 4. Find the slope at the point (, 0) of the curve parameterized by x(t) = cos t + t, y(t) = t 4 + t. Solution. We have dy dx = ẏ ẋ = 4t3 + sin t + We investigate to which parameter value the given point corresponds. The y- coordinate gives t(t 3 + ) = 0, so t = 0 or t =. For t = 0 we get the correct x-coordinate, for t = we don t. So t = 0 is the parameter value. So the slope in the given point equals sin 0 + =..7 Implicit differentiation Curves sometimes come in the form of an equation relation x and y from which y cannot (easily) be solved. It is however still possible to compute the slope dy dx at a given point on the curve through what is called implicit differentiation. A formal formulation of this requires differentiation of functions of several variables (which we will do only later), but on examples it is not to difficult too figure out what to do. We illustrate the method on the simple example of the unit circle. So we have the equation x + y = and we want to know that dy dx is. What we do is differentiate both sides of the equation with respect to x (that of course gives another identity) and keep in mind that y is a function of x. For the unit circle that gives: d dx (x + y ) = d dx, and taking the derivatives gives (using dy dx This equation can be solved for dy dx as = dy dy x + y dy dx = 0. dy dx = x y. dy dx ): Using that for the upper half of the unit circle we have y = x and for the lower half y = x it can be checked that this is in both cases the right formula for the slope. Problem 43. Calculate the slope at the point p = ( 3 π, π ) of the curve x3 y + sin y =. 9

31 Solution. Implicit differentiation gives Substituting the point gives 3 3x y + x 3 dy dx + dy cos y = 0. dx ( ) /3 π π + π dy dx dy (p) + (p) 0 = 0. dx So the slope is 3 ( ) /3 (π ) ( π ) 4/3 = 3. π.8 Maxima and minima An important application of derivatives is that it allows us to find maxima and minima of functions. But before we go into computing these things lets first define them properly and say something about when they exist. The function f has a local maximum at x = c if f(c) f(x) for all point x close to c. It is a global maximum if this holds for all x for which f is defined. The function f has a local minimum at x = c if f(c) f(x) for all point x close to c. It is a global minimum if this holds for all x for which f is defined. Extremum is another word for something that is either a maximum or a minimum. The following existence results is very difficult to prove: a continuous function defined on a finite interval that includes the endpoints has a global maximum and a global minimum. Note that the extremum may be reached at the endpoints, a simple example is f(x) = x when considered on the interval 0 x ; the minimum is zero and is reached at zero and the maximum is and is reached at. If we were to exclude the endpoints, i.e. consider f(x) = x on the interval 0 < x <, then the function has no maximum and no minimum. The following result is of great use in finding extrema: if f has a local extremum at x = c and f is differentiable in c, then f (c) = 0. Of course a global extremum is also a local extremum, so this procedure helps us to find global extrema as well. To find local extrema of f we do the following:. calculate the derivative,. solve f (x) = 0, 3. find points where f is not defined, 4. consider the endpoints of the interval on which f is defined. 30

32 We already showed that step 4 should not be forgotten when we considered the example f(x) = x on the interval [0, ]. We now show that step 3 should not be forgotten. Consider the absolute value function (see figure.). This function obviously has its minimum in x = 0, but the derivative is for x < 0, it is for x > 0 and undefined for x = 0. So solving f (x) = 0 doesn t provide you with the minimum. We further note that the points found in steps,3 and 4 are only candidates for local extrema, they may not be local extrema. An example here is f(x) = x 3 on the interval [, ] for which f (0) = 0, but there is no local extremum in x = 0 (see figure.4). We will say a bit more about this later Figure.4: The graph of x 3 Problem 44. Find the global maximum and global minimum of f(x) = x on the interval [, ]. Solution. Of course for this example the maximum and minimum are easily seen, but we do the procedure anyway. We have f (x) = x, so f (x) = 0 if and only if x = 0. The value at this point is 0. There are no point where f is not differentiable. At the boundary point x = the value is, at the boundary point x = the value is 4. So the global minimum value is 0 and is reached in x = 0, the global maximum is 4 and is reached in x =. Determining whether a point is a local minimum, a local maximum or neither is a bit, but not much, more difficult than determining whether it is a global minimum, a global maximum or neither. Now we do not compare all values at the points found in steps,3 and 4 with each other, but only neighboring ones. We illustrate this on the above example. Problem 45. Determine the local extrema and their nature of f(x) = x on the interval [, ] using the evaluation method. Solution. As shown in problem 44 the points to be considered are x =, x = 0 and x =. The values can be given conveniently in the following chart

33 The value at x = is higher than that at neighboring special points (x = 0), so at x = there is a local maximum with value. The value at x = 0 is lower than that at the neighboring special points (x = and x = ), so at x = 0 there is a local minimum with value 0. The value at x = is higher than that at neighboring special points (x = 0), so at x = there is a maximum with value 4. In the above problem we checked whether the points we found in steps,3 and 4 of the procedure were maxima or minima by simply substituting them into the function and comparing values. There are two ways other of checking whether a point is a local maximum, a local minimum or neither. These two other ways have the advantage that they generalize to functions of several variables, whereas the above used method does not (the problem is that the concept of neighboring point doesn t generalize). The first is by looking at the sign of the derivative. The sign of the derivative tells us whether the function is increasing (positive derivative) or decreasing (negative derivative). There is a theorem that says that the derivative can only change sign at points where it goes through zero or points where it doesn t exist (these points we found in steps and 3). We can find the sign of the derivative in between such points by e.g. substituting a particular point in that interval into the derivative and looking at the sign of the outcome. The change of sign of the derivative gives us the following information. Suppose that f is continuous at c and that c is an interior point if the derivative changes sign from positive to negative at x = c, then f has a local maximum in c, if the derivative changes sign from negative to positive at x = c, then f has a local minimum in c, if the derivative does not change sign, then f does not have a local extremum at x = c. If x = c is an endpoint, then the derivative can obviously not change sign since it is not defined on one side of x = c. We can nonetheless apply the above result pretending that the derivative changes sign at the endpoint. We will illustrate this on the already discussed example. Problem 46. Determine the local extrema and their nature of f(x) = x on the interval [, ] using the change of sign of the derivative method. Solution. As shown in problem 44 the points to be considered are x =, x = 0 and x =. The derivative is f (x) = x and its sign chart is + 0 At the endpoint x = the sign changes from positive to negative (using our pretense that the sign changes at endpoints), so at x = there is a local 3

34 maximum with value. At x = 0 the sign changes from negative to positive, so at x = 0 there is a local minimum with value 0. At the endpoint x = the sign changes from positive to negative (again using our pretense that the sign changes at endpoints), so at x = there is a local maximum with value 4. The third method of determining if we are dealing with a local maximum, a local minimum or neither is by considering higher order derivatives. Suppose that f (c) = 0 if f (c) > 0, then f has a local minimum in x = c, if f (c) < 0, then f has a local maximum in x = c, if f (c) = 0, then f might have a local maximum, a local minimum or neither in x = c. An example where f (0) = f (0) = 0 and f has a local minimum in x = 0 is f(x) = x 4, an example where it has a local maximum is f(x) = x 4 and an example where it has neither is f(x) = x 3. We will look deeper into this issue when we consider Taylor polynomials. Note that this method of looking at the second derivative says nothing in the case of points obtained in steps 3 or 4 (points where f does not exist and endpoints), it really is crucial that f (c) = 0. Problem 47. Determine the local extrema and their nature of f(x) = x on the interval [, ] using the method of higher order derivatives. Solution. As shown in problem 44 the points to be considered are x =, x = 0 and x =. The second derivative is the constant. So there is a local minimum at x = 0 with value 0. Since x = 0 is a local minimum, it follows that the endpoints x = and x = are local maxima. We now illustrate that the second derivative really does not say anything about the nature of local extrema if f (c) 0. We first consider the case where the derivative does not exist. The examples are the absolute value function x + x and g(x) = x + x. The second derivatives of f and g are the same; they are everywhere, except in x = 0 where they are undefined. However, f has a minimum in x = 0, whereas g has a maximum in x = 0. So even though the second derivatives are identical, the nature of extrema is different. We now consider the case of endpoints. Consider the functions f(x) = x + x and g(x) = x + x on the interval [0, ]. The second derivative is identical to in both cases. However, f has a minimum in x = 0 and a maximum in x = and for g it is exactly the other way around. So again, the second derivatives are identical, but the nature of the extrema is different. So f (c) = 0 is really crucial for the method of higher order derivatives to work! Problem 48. Find the location of the local extrema of f(x) = x e x on the interval [ 3, ] and classify them as either local maximum or local minimum. 33