OpenStax-CNX moule: m39313 1 Differential Calculus: Differentiation (First Principles, Rules) an Sketcing Graps (Grae 12) * Free Hig Scool Science Texts Project Tis work is prouce by OpenStax-CNX an license uner te Creative Commons Attribution License 3.0 1 Dierentiation from First Principles Te tangent problem as given rise to te branc of calculus calle ierential calculus an te equation: f(x+) f(x) lim 0 enes te erivative of te function f (x). Using to calculate te erivative is calle ning te erivative from rst principles. Denition 1: Derivative Te erivative of a function f (x) is written as f ' (x) an is ene by: f ' (x) = lim 0 f (x + ) f (x) Tere are a few ierent notations use to refer to erivatives. If we use te traitional notation y = f (x) to inicate tat te epenent variable is y an te inepenent variable is x, ten some common alternative notations for te erivative are as follows: f ' (x) = y ' = y = f = f (x) = Df (x) = D xf (x) Te symbols D an are calle ierential operators because tey inicate te operation of ierentiation, wic is te process of calculating a erivative. It is very important tat you learn to ientify tese ierent ways of enoting te erivative, an tat you are consistent in your usage of tem wen answering questions. tip: Toug we coose to use a fractional form of representation, y is a limit an is not a fraction, y i.e. oes not mean y. y means y ierentiate wit respect to x. Tus, p means p ierentiate wit respect to x. Te ` ' is te operator", operating on some function of x. * Version 1.1: Aug 1, 2011 3:01 am -0500 ttp://creativecommons.org/licenses/by/3.0/
OpenStax-CNX moule: m39313 2 Vieo on calculus - 2 Tis meia object is a Flas object. Please view or ownloa it at <ttp://www.youtube.com/v/gpmgmca2sby?version=3> Exercise 1: Derivatives - First Principles (Solution on p. 7.) Calculate te erivative of g (x) = x 1 from rst principles. 1.1 Derivatives 1. Given g (x) = x 2 a. etermine g(x+) g(x) g(x+) g(x) b. ence, etermine lim 0 c. explain te meaning of your answer in (b). 2. Fin te erivative of f (x) = 2x 2 + 3x using rst principles. 3. Determine te erivative of f (x) = 1 x 2 using rst principles. 4. Determine f ' (3) from rst principles if f (x) = 5x 2. 5. If (x) = 4x 2 4x, etermine ' (x) using rst principles. 2 Rules of Dierentiation Calculating te erivative of a function from rst principles is very long, an it is easy to make mistakes. Fortunately, tere are rules wic make calculating te erivative simple. 2.1 Investigation : Rules of Dierentiation From rst principles, etermine te erivatives of te following: 1. f (x) = b 2. f (x) = x 3. f (x) = x 2 4. f (x) = x 3 5. f (x) = 1/x You soul ave foun te following: f (x) f ' (x) b 0 x 1 x 2 2x x 3 3x 2 1/x = x 1 x 2
OpenStax-CNX moule: m39313 3 Table 1 If we examine tese results we see tat tere is a pattern, wic can be summarise by: (xn ) = nx n 1 Tere are two oter rules wic make ierentiation simpler. For any two functions f (x) an g (x): [f (x) ± g (x)] = f ' (x) ± g ' (x) Tis means tat we ierentiate eac term separately. Te nal rule applies to a function f (x) tat is multiplie by a constant k. [k.f (x)] = kf ' (x) Vieo on calculus - 3 Tis meia object is a Flas object. Please view or ownloa it at <ttp://www.youtube.com/v/sjecbgk-sny?version=3> Exercise 2: Rules of Dierentiation (Solution on p. 7.) Determine te erivative of x 1 using te rules of ierentiation. 2.2 Summary of Dierentiation Rules Given two functions, f (x) an g (X) we know tat: b = 0 (xn ) = nx n 1 f (kf) = k f (f + g) = + g Table 2 2.2.1 Rules of Dierentiation 1. Fin f ' (x) if f (x) = x2 5x+6 x 2. 2. Fin f ' (y) if f (y) = y. 3. Fin f ' (z) if f (z) = (z 1) (z + 1). 4. Determine y if y = x3 +2 x 3 x. 5. Determine te erivative of y = x 3 + 1 3x. 3
OpenStax-CNX moule: m39313 4 3 Applying Dierentiation to Draw Graps Tus far we ave learnt about ow to ierentiate various functions, but I am sure tat you are beginning to ask, Wat is te point of learning about erivatives? Well, we know one important fact about a erivative: it is a graient. So, any problems involving te calculations of graients or rates of cange can use erivatives. One simple application is to raw graps of functions by rstly etermine te graients of straigt lines an seconly to etermine te turning points of te grap. 3.1 Fining Equations of Tangents to Curves In "Average Graient an Graient at a Point" we saw tat ning te graient of a tangent to a curve is te same as ning te graient (or slope) of te same curve at te point of te tangent. We also saw tat te graient of a function at a point is just its erivative. Since we ave te graient of te tangent an te point on te curve troug wic te tangent passes, we can n te equation of te tangent. Exercise 3: Fining te Equation of a Tangent to a Curve (Solution on p. 7.) Fin te equation of te tangent to te curve y = x 2 at te point (1,1) an raw bot functions. 3.2 Curve Sketcing Dierentiation can be use to sketc te graps of functions, by elping etermine te turning points. We know tat if a grap is increasing on an interval an reaces a turning point, ten te grap will start ecreasing after te turning point. Te turning point is also known as a stationary point because te graient at a turning point is 0. We can ten use tis information to calculate turning points, by calculating te points at wic te erivative of a function is 0. tip: If x = a is a turning point of f (x), ten: f ' (a) = 0 Tis means tat te erivative is 0 at a turning point. Take te grap of y = x 2 as an example. We know tat te grap of tis function as a turning point at (0,0), but we can use te erivative of te function: y ' = 2x an set it equal to 0 to n te x-value for wic te grap as a turning point. 2x = 0 x = 0 We ten substitute tis into te equation of te grap (i.e. y = x 2 ) to etermine te y-coorinate of te turning point: f (0) = (0) 2 = 0 Tis correspons to te point tat we ave previously calculate. Exercise 4: Calculation of Turning Points (Solution on p. 7.) Calculate te turning points of te grap of te function f (x) = 2x 3 9x 2 + 12x 15. We are now reay to sketc graps of functions. Meto: Sketcing Graps: Suppose we are given tat f (x) = ax 3 + bx 2 + cx +, ten tere are ve steps to be followe to sketc te grap of te function: 1. If a > 0, ten te grap is increasing from left to rigt, an may ave a maximum an a minimum. As x increases, so oes f (x). If a < 0, ten te grap ecreasing is from left to rigt, an as rst a minimum an ten a maximum. f (x) ecreases as x increases. 2. Determine te value of te y-intercept by substituting x = 0 into f (x)
OpenStax-CNX moule: m39313 5 3. Determine te x-intercepts by factorising ax 3 + bx 2 + cx + = 0 an solving for x. First try to eliminate constant common factors, an to group like terms togeter so tat te expression is expresse as economically as possible. Use te factor teorem if necessary. 4. Fin te turning points of te function by working out te erivative f an setting it to zero, an solving for x. 5. Determine te y-coorinates of te turning points by substituting te x values obtaine in te previous step, into te expression for f (x). 6. Use te information you're given to plot te points an get a roug iea of te graients between points. Ten ll in te missing parts of te function in a smoot, continuous curve. Exercise 5: Sketcing Graps (Solution on p. 8.) Draw te grap of g (x) = x 2 x + 2 Exercise 6: Sketcing Graps (Solution on p. 9.) Sketc te grap of g (x) = x 3 + 6x 2 9x + 4. 3.2.1 Sketcing Graps 1. Given f (x) = x 3 + x 2 5x + 3: a. Sow tat (x 1) is a factor of f (x) an ence fatorise f (x) fully. b. Fin te coorinates of te intercepts wit te axes an te turning points an sketc te grap 2. Sketc te grap of f (x) = x 3 4x 2 11x + 30 sowing all te relative turning points an intercepts wit te axes. 3. a. Sketc te grap of f (x) = x 3 9x 2 + 24x 20, sowing all intercepts wit te axes an turning points. b. Fin te equation of te tangent to f (x) at x = 4. 3.3 Local minimum, Local maximum an Point of Inextion If te erivative ( y ) is zero at a point, te graient of te tangent at tat point is zero. It means tat a turning point occurs as seen in te previous example. From te rawing te point (1;0) represents a local minimum an te point (3;4) te local maximum. A grap as a orizontal point of inexion were te erivative is zero but te sign of te sign of te graient oes not cange. Tat means te grap always increases or always ecreases.
OpenStax-CNX moule: m39313 6 From tis rawing, te point (3;1) is a orizontal point of inexion, because te sign of te erivative oes not cange from positive to negative.
OpenStax-CNX moule: m39313 7 Solutions to Exercises in tis Moule Solution to Exercise (p. 2) Step 1. We know tat te graient at a point x is given by: g ' g(x+) g(x) (x) = lim 0 Step 2. g (x + ) = x + 1 Step 3. g ' g(x+) g(x) (x) = lim 0 x+ 1 (x 1) = lim 0 x+ 1 x+1 = lim 0 = lim 0 = lim1 0 = 1 Step 4. Te erivative g ' (x) of g (x) = x 1 is 1. Solution to Exercise (p. 3) Step 1. We will apply two rules of ierentiation: (xn ) = nx n 1 an [f (x) g (x)] = [f (x)] [g (x)] Step 2. In our case f (x) = x an g (x) = 1. f ' (x) = 1 an g ' (x) = 0 Step 3. Te erivative of x 1 is 1 wic is te same result as was obtaine earlier, from rst principles. Solution to Exercise (p. 4) Step 1. We are require to etermine te equation of te tangent to te curve ene by y = x 2 at te point (1,1). Te tangent is a straigt line an we can n te equation by using erivatives to n te graient of te straigt line. Ten we will ave te graient an one point on te line, so we can n te equation using: y y 1 = m (x x 1 ) from grae 11 Coorinate Geometry. Step 2. Using our rules of ierentiation we get: y ' = 2x Step 3. In orer to etermine te graient at te point (1,1), we substitute te x-value into te equation for te erivative. So, y ' at x = 1 is: m = 2 = 2 Step 4. y y 1 = m (x x 1 ) y 1 = (2) (x 1) y = 2x 2 + 1 y = 2x 1 Step 5. Te equation of te tangent to te curve ene by y = x 2 at te point (1,1) is y = 2x 1. Step 6. Solution to Exercise (p. 4) Step 1. Using te rules of ierentiation we get: f ' (x) = 6x 2 18x + 12
OpenStax-CNX moule: m39313 8 Step 2. Step 3. 6x 2 18x + 12 = 0 x 2 3x + 2 = 0 (x 2) (x 1) = 0 Terefore, te turning points are at x = 2 an x = 1. f (2) = 2(2) 3 9(2) 2 + 12 (2) 15 = 16 36 + 24 15 = 11 f = 2 3 9 2 + 12 15 = 2 9 + 12 15 = 10 Step 4. Te turning points of te grap of f (x) = 2x 3 9x 2 + 12x 15 are (2,-11) an (1,-10). Solution to Exercise (p. 5) Step 1. Te y-intercept is obtaine by setting x = 0. g (0) = (0) 2 0 + 2 = 2 Te turning point is at (0,2). Step 2. Te x-intercepts are foun by setting g (x) = 0. g (x) = x 2 x + 2 0 = x 2 x + 2 Using te quaratic formula an looking at b 2 4ac we can see tat tis woul be negative an so tis function oes not ave real roots. Terefore, te grap of g (x) oes not ave any x-intercepts. Step 3. Work out te erivative g an set it to zero to for te x coorinate of te turning point. g = 2x 1 g = 0 2x 1 = 0 2x = 1 x = 1 2 Step 4. y coorinate of turning point is given by calculating g ( 1 2). Te turning point is at ( 1 2, ) 7 4 g ( ) 1 2 = ( 1 2) 2 ( 1 2) + 2 = 1 4 1 2 + 2 = 7 4 Step 5.
OpenStax-CNX moule: m39313 9 Solution to Exercise (p. 5) Step 1. Fin te turning points by setting g ' (x) = 0. If we use te rules of ierentiation we get g ' (x) = 3x 2 + 12x 9 g ' (x) = 0 3x 2 + 12x 9 = 0 x 2 4x + 3 = 0 (x 3) (x 1) = 0 Te x-coorinates of te turning points are: x = 1 an x = 3. Te y-coorinates of te turning points are calculate as: g (x) = x 3 + 6x 2 9x + 4 g = 3 + 6 2 9 + 4 = 1 + 6 9 + 4 = 0 g (x) = x 3 + 6x 2 9x + 4 g (3) = (3) 3 + 6(3) 2 9 (3) + 4 = 27 + 54 27 + 4 = 4 Terefore te turning points are: (1, 0) an (3, 4). Step 2. We n te y-intercepts by ning te value for g (0). g (x) = x 3 + 6x 2 9x + 4 y int = g (0) = (0) 3 + 6(0) 2 9 (0) + 4 = 4 Step 3. We n te x-intercepts by ning te points for wic te function g (x) = 0. g (x) = x 3 + 6x 2 9x + 4 Use te factor teorem to conrm tat (x 1) is a factor. If g = 0, ten (x 1) is a factor. g (x) = x 3 + 6x 2 9x + 4 g = 3 + 6 2 9 + 4 = 1 + 6 9 + 4 = 0 Terefore, (x 1) is a factor. If we ivie g (x) by (x 1) we are left wit: x 2 + 5x 4 Tis as factors (x 4) (x 1) Terefore: g (x) = (x 1) (x 1) (x 4) Te x-intercepts are: x int = 1, 4
OpenStax-CNX moule: m39313 10 Step 4.