The rules of calculus. Christopher Thomas

Transcription

1 Mathematics Learning Centre The rules of calculus Christopher Thomas c 1997 University of Syney

2 Mathematics Learning Centre, University of Syney 1 1 How o we fin erivatives (in practice)? Differential calculus is a proceure for fining the exact erivative irectly from the formula of the function, without having to use graphical methos. In practise we use a few rules that tell us how to fin the erivative of almost any function that we are likely to encounter. In this section we will introuce these rules to you, show you what they mean an how to use them. Warning! To follow the rest of these notes you will nee feel comfortable manipulating expressions containing inices. If you fin that you nee to revise this topic you may fin the Mathematics Learning Centre publication Exponents an Logarithms helpful. 1.1 Derivatives of constant functions an powers Perhaps the simplest functions in mathematics are the constant functions an the functions of the form x n. Rule 1 If k is a constant then x k =0. Rule If n is any number then x xn = nx n 1. Rule 1 at least makes sense. The graph of a constant function is a horizontal line an a horizontal line has slope zero. The erivative measures the slope of the tangent, an so the erivative is zero. How you approach Rule is up to you. You certainly nee to know it an be able to use it. However we have given no justification for why Rule works! In fact in these notes we will give little justification for any of the rules of ifferentiation that are presente. We will show you how to apply these rules an what you can o with them, but we will not make any attempt to prove any of them. Examples If f(x) =x 7 then f (x) =7x 6. If y = x 0.5 then y x = 0.5x 1.5. x x 3 = 3x 4. If g(x) =3. then g (x) =0. If f(t) =t 1 then f (t) = 1 t 1. If h(u) = 13.9 then h (u) =0.

3 Mathematics Learning Centre, University of Syney In the examples above we have use Rules 1 an to calculate the erivatives of many simple functions. However we must not lose sight of what it is that we are calculating here. The erivative gives the slope of the tangent to the graph of the function. For example, if f(x) =x then f (x) =x. Tofin the slope of the tangent to the graph of x at x =1we substitute x =1into the erivative. The slope is f (1) = 1=. Similarly the slope of the tangent to the graph of x at x = 0.5 isfoun by substituting x = 0.5 into the erivative. The slope is f ( 0.5) = 0.5 = 1. This is illustrate in Figure slope = f ' (-0.5) = (-0.5, 0.5) ( 1, 1 ) slope = f ' (1) = Example Figure 1: Slopes of tangents to the graph of y = x. Fin the slope of the tangent to the graph of the function g(t) =t 4 at the point on the graph where t =. Solution The erivative is g (t) =4t 3, an so the slope of the tangent line at t = isg ( ) = 4 ( ) 3 = 3. Example Fin the equation of the line tangent to the graph of y = f(x) =x 1 at the point x =4. Solution f(4) = 4 1 = 4 =,sothe coorinates of the point on the graph are (4, ). The erivative is f (x) = x 1 = 1 x an so the slope of the tangent line at x =4isf (4) = 1.Wetherefore know the slope of 4 the line an we know one point through which the line passes. Any non vertical line has equation of the form y = mx + b where m is the slope an b the vertical intercept.

4 Mathematics Learning Centre, University of Syney 3 In this case the slope is 1,som = 1, an the equation is y = x + b. Because the line passes through the point (4, ) we know that y =when x =4. Substituting we get = b, sothat b =1. The equation is therefore y = x Notice that in the examples above the inepenent variable is not always calle x. We have also use u an t, an in fact we can an will use many ifferent letters for the inepenent variable. Notice also that we might not stick to the symbol f to stan for function. Many other symbols are use. Some of the common ones are g an h. Throughout this booklet we will use a variety of symbols for functions an variables to get you use to the fact that our choice of symbols makes no ifference to the ieas that we are introucing. On the other han, we can make life easier for ourselves if we make sensible choices of symbols. For example if we were iscussing the revenue obtaine by a manufacturer who sells articles for a certain price it might be sensible for us to choose the symbol p to mean price, an r to mean revenue, an to write r(p) toexpress the fact that the revenue is a function of the price. In this way the symbols we have chosen remin us of their meaning, much more than if we ha chosen x to represent price an f to represent revenue an written f(x). On the other han, because the symbol has a f (x) special use in calculus, to express the erivative,wealmost never use for any other x purpose. For this reason you will often see the letter s use to represent isplacement. We now know how to ifferentiate any function that is a power of the variable. Examples are functions like x 3 an t 1.3.You will come across functions that o not at first appear to be a power of the variable, but can be rewritten in this form. One of the simplest examples is the function f(t) = t, which can also be written in the form f(t) =t 1. The erivative is then f (t) = t 1 = 1 t. Similarly, if then Examples h(s) = 1 s = s 1 h (s) = s = 1 s. If f(x) = 1 3 x = x 1 3 then f (x) = 1 3 x 4 3. If y = 1 x x = 3 y x then x = 3 x 5.

5 Mathematics Learning Centre, University of Syney 4 Exercises 1.1 Differentiate the following functions: a. f(x) =x 4 b. y = x 7 c. f(u) =u.3. f(t) =t 1 3 e. f(t) =t 7 f. g(z) =z 3 g. y = t 3.8 h. z = x 3 7 Exercise 1. Express the following as powers an then ifferentiate: 1 a. b. t t c. 3 x x 1. x x e. 1 x 4 x f. s 3 s 3 s 1 t g. h. u 3 t i. x 1 x t x Exercise 1.3 Fin the equation of the line tangent to the graph of y = 3 x when x =8. 1. Aing, subtracting, an multiplying by a constant So far we know how to ifferentiate powers of the inepenent variable. Many of the functions that you will encounter are mae up in simple ways from powers. For example, a function like 3x is just a constant multiple of x. However neither Rule 1 nor Rule tell us how to ifferentiate 3x. Nor o they tell us how to ifferentiate something like x + x 3 or x x 3. Rules 3 an 4 specify how to ifferentiate combinations of functions that are forme by multiplying by constants, or by aing or subtracting functions. Rule 3 If f(x) =cg(x), where c is a constant, then f (x) =cg (x). Rule 4 If f(x) =g(x) ± h(x) then f (x) =g (x) ± h (x). Examples If f(x) =3x then f (x) =3 x x =6x. If g(t) =3t +t then g (t) = t 3t + t t =6t 4t 3. If y = 3 x x 3 x =3x 1 x 4 3 then y x = 3 x x 1 3. If y = 0.3x 0.4 then y x =0.1x 1.4. x x0.3 =0.6x 0.7.

6 Mathematics Learning Centre, University of Syney 5 Warning! Although Rule 4 tells us that x (f(x) ± g(x)) = f (x) ± g (x), the same is not true for multiplication or ivision. To ifferentiate f(x) g(x) or f(x) g(x) we cannot simply fin f (x) an g (x) an multiply or ivie them. Be careful of this! The methos of ifferentiating proucts of functions or quotients of functions are iscusse in Sections 1.3 an 1.4. Exercise 1.4 Differentiate the following functions: a. f(x) =5x x b. y =x x c. f(t) =.5t.3 + t t. h(z) =z z e. f(u) =u 5 3 3u 7 f. g(z) =8z 5 z g. y =5t 8 + t t h. z =4x 1 7 +x The prouct rule Another way of combining functions to make new functions is by multiplying them together, or in other wors by forming proucts. The prouct rule tells us how to ifferentiate functions like this. Rule 5 (The prouct rule) If f(x) =u(x)v(x) then f (x) =u(x)v (x)+u (x)v(x). Examples If y =(x + )(x +3)then y =(x + )x +1(x +3). If f(x) = x(x 3 3x +7)then f (x) = x(3x 6x)+ 1 x 1 (x 3 3x + 7). If z =(t + 3)( t + t 3 ) then z t =(t + 3)( 1 t 1 +3t )+t( t + t 3 ). Exercise 1.5 Use the prouct rule to ifferentiate the functions below:

7 Mathematics Learning Centre, University of Syney 6 a. f(x) =(4x 3 + )(1 3x) b. g(x) =(x + x + )(x +1) c. h(x) =(3x 3 x +8x 5)(x x +4). f(s) =(1 1 s )(3s +5) e. g(t) =( t + 1 )(t 1) t f. h(y) =( y + y )(1 3y ) Exercise 1.6 If r =(t + 1 t )(t t + 1), fin the rate of change of r with respect to t when t =. Exercise 1.7 Fin the slope of the tangent to the curve y =(x x + 1)(3x 3 5x +)atx =. 1.4 The Quotient Rule This rule allows us to ifferentiate functions which are forme by iviing one function by another, ie by forming quotients of functions. An example is such as x +3 f(x) = 3x 5. Rule 6 (The quotient rule) f(x) = u(x) v(x) f (x) = v(x)u (x) u(x)v (x) [v(x)] = vu uv v. Warning! Because of the minus sign in the numerator (ie in the top line) it is important to get the terms in the numerator in the correct orer. This is often a source of mistakes, so be careful. Decie on your own way of remembering the correct orer of the terms.

8 Mathematics Learning Centre, University of Syney 7 Examples If y = x +3x y, then x 3 +1 x = (x3 + 1)(4x +3) (x +3x)3x. (x 3 +1) If g(t) = t +3t +1 t +1 then g (t) = ( t + 1)(t +3) (t +3t + 1)( 1 t 1 ) ( t +1). Exercise 1.8 Use the Quotient Rule to fin erivatives for the following functions: a. f(x) = x 1 x +1 c. h(x) = x + x +5 e. f(s) = 1+ s 1 s g. f(u) = u3 + u 4 3u 4 +5 b. g(x) =. f(t) = x +3 3x t 1+t f. h(x) = x 1 x 3 +4 h. g(t) = t(t +6) t +3t Solutions to exercises Exercise 1.1 a. f (x) =4x 3 b. y x = 7x 8 c. f (u) =.3u 1.3. f (t) = 1 3 t 4 3 e. f (t) = 7 t 15 7 f. g (z) = 3 z 5 g. y t = 3.8t 4.8 h. z x = x 7

9 Mathematics Learning Centre, University of Syney 8 Exercise 1. a. c. e. g. i. 1 x = x so ( ) 1 = x 3 b. t t = t 3 x x so ( ) 3 t t = t t 1 ( ) 1 x x 3 x = x x = ( ) 1 x 4 x 3 x = x x5/4 = x 4 f. ( ) 1 = u 3 u u 3 u = 3u 4 h. ( ) x x 1/ = x x x 1=0 x x t = 5 x x = 5 x x ( ) s 3 s 3 = s s s = 6 s 13 6 ( ) t t = 3 3 t t t = x 5 t 7 Exercise 1.3 When x =8wehave y = 3 8=sothe point (8, ) is on the line. Now y when x =8. The tangent therefore has equation y x = 1 1 x = x 3 3 an so y = 1 1 x + b. Substituting x = 8 an y = into this equation we obtain = b so that b = 4 3. The equation is therefore y = x Exercise 1.4 a. f (x) =10x x 1 b. y x = 14x 8 6x 3 c. f (t) =5.75t t 1. h (z) = 1 3 z e. f (u) = 5 3 u/3 +1u 8 f. g (z) = 16z 3 +5z g. y t = 40t t 1 h. z x = 4 7 x 6 7 x 3

10 Mathematics Learning Centre, University of Syney 9 Exercise 1.5 a. f (x) =1x (1 3x) 3(4x 3 +) b. g (x) =(x + 1)(x +1)+(x + x + )x c. h (x) =(9x 4x + 8)(x x +4)+(3x 3 x +8x 5)(x ). f (s) = s(3s +5)+3(1 s ) e. g (t) =( t 1 t )(t 1)+( t + 1 t ) f. h (y) =( 1 y +y)(1 3y ) 6y( y + y ) Exercise 1.6 The rate of change of r with respect to t is r t =(1 t )(t t +1)+(t + 1 )(t ). t Substituting t =we obtain (1 1)(4 4+1)+(+ 1)(4 ) = Exercise 1.7 The graient of the tangent is given by y x =(x )(3x3 5x +)+(x x + 1)(9x 10x). Substituting x = we obtain 8. Exercise 1.8 a. f (x) = b. g (x) = (x +1) (x 1) (x +1) = (x +1) (3x ) (x + 3)3 (3x ) = 13 (3x ) c. h (x) = (x + 5)x (x + )x (x +5) =. f (t) = (1 + t ) 8t (1+t ) = 4t (1+t ) 6x (x +5) e. f (s) = (1 s) 1 s 1 +(1+ s) 1 s 1/ (1 = s) s 1 (1 s) f. h (x) = (x3 + 4)x (x 1)3x = x4 +3x +8x (x 3 +4) (x 3 +4) g. f (u) = (3u4 + 5)(3u +1) (u 3 + u 4)1u 3 = 3u6 9u 4 +48u 3 +15u +5 (3u 4 +5) (3u 4 +5) h. g (t) = (t +3t + 1)(t +6) (t +6t)(t +3) = 3t +t +6 (t +3t +1) (t +3t +1)