Differentiation Techniques

Transcription

1 C H A P T E R Differentiation Techniques Objectives To differentiate functions having negative integer powers. To understand and use the chain rule. To differentiate rational powers. To find second derivatives of functions. To use limits to define continuit at a point. To use differentiation techniques to sketch graphs of functions. To appl differentiation techniques to solving problems. The work in this chapter is not part of the content for VCE Mathematical Methods (CAS) Units&.Itisincluded as a useful foundation for VCE Mathematical Methods (CAS) Units3&4.. Differentiating n where n is a negative integer In earlier chapters the differentiation of polnomial functions has been introduced and applications of differentiation of such functions have been considered. In this section we add to the famil of functions for which we can find the derived functions. In particular, we will consider functions which involve linear combinations of powers of where a power ma be a negative integer. e.g. f : R \{0} R, f () = f : R \{0} R, f () = + f : R \{0} R, f () = Note: We have reintroduced the function notation which emphasises the need for consideration of domain. 58

2 Chapter Differentiation Techniques 583 Eample Let f : R \{0} R, f () =.Find f () bfirst principles. 0 P, Q + h, + h f ( + h) f () Gradient of chord PQ = + h = + h h ( + h) = ( + h) h = h ( + h) h = ( + h) The gradient of the curve at P = lim h 0 ( + h) = =, i.e. f () = Eample Let f : R \{0} R, f () = 3.Find f () bfirst principles. 0 P(, 3 ) Gradient of chord PQ = ( + h) 3 3 Q ( + h, ( + h) 3 ) h = 3 ( + h) 3 ( + h) 3 3 h = 3 ( h + 3h + h 3 ) ( + h) 3 3 h = 3 h 3h h 3 ( + h) 3 3 h = 3 3h h ( + h) 3 3 The gradient of the curve at P = lim h 0 3 3h h ( + h) 3 3 = 3 6 = 3 4 = 3 4 i.e. f () = 3 4

3 584 Essential Mathematical Methods &CAS We are now in a position to state the generalisation of the result we found in Chapter 9. For f () = n, f () = n n, n is a non-zero integer. For f () = c, f () = 0, where c is a constant. We note that for n wetake the domain of f to be R \ {0}, and for n wetake the domain of f to be R. Eample 3 Find the derivative of , 0. If f () = , 0 f () = 4 3 ( 3 4 ) + ( ) + (0) = , 0 Eample 4 Find the derivative f of f : R \ {0} R, f () = f : R \{0} R, f () = 3() 6( 3 ) + (0) = Eample 5 Find the gradient of the curve determined b the function f : R \{0}, f () = + at the point (, ). f : R \{0} R, f () = + ( ) = and f () = = The gradient of the curve is at the point (, ). Eample 6 Show that the gradient of the function f : R \{0} R, f () = 3 is alwas negative.

4 f : R \{0} R, f () = 3 4 = is positive for all and thus f () < 0 for all 0. Eercise A Chapter Differentiation Techniques 585 Eamples 3, 4 Eamples, Eample 5 Differentiate each of the following with respect to : a b + 5 d e f Find the derivative of each of the following: a 3z + z + 4 z, z 0 b 3 + z z 3, z 0 c d 9z + 4z + 6z 3, z 0 e 9 z, z 0 f 3 a Carefull sketch the graph of f () =, 0. b c z + 3z, z 0 4z 5z 3z, z 0 5z Let P be the point (, f ()) and Q the point ( + h, f ( + h)). Find the gradient of the chord PQ. c Hence find the gradient of the curve f () = at =. d Find the equation of the normal to the curve with equation = at the point (, ). 4 Find the gradient of each of the following curves at the given point: a = + 3, 0, at ( ), 8 b =, n 0, at ( ) 4, 4 c =, 0, at (, 0) d = ( + 3 ), 0, at (, ). 5 Find the -coordinate of the point on the curve with equation f () = and gradient: a 6 b 6. The chain rule An epression such as f () = ( 3 + ) ma be differentiated b epanding and then differentiating each term. This method is a great deal more tiresome for an epression such as f () = ( 3 + ) 30.Wetransform f () = ( 3 + ) into two simpler functions defined b: h() = 3 + (= u) and g(u) = u (= ) which are chained together: h u g

5 586 Essential Mathematical Methods &CAS We ma eploit this connection to differentiate. We use Leibniz notation to eplore this idea. Consider f () = ( 3 + ) at the point where =. u u = 3 + = u R P Q δ δu S δu δ u (0, ) Z W δ 0 = ( 3 + ) δ When =, u = 9 When u = 9, = 8 When =, = 8 P is the point (, 9) R is the point (9, 8) Z is the point (, 8) A section of a spreadsheet (below) illustrates the connection between the gradients of the chords PQ, RS and WZ.Itcan be seen that as gets smaller so does u and = u u. B considering u becoming smaller and hence becoming smaller, it is seen that d d = d du du d This is called the chain rule for differentiation. u u u u u u From the spreadsheet it can be seen that the gradient of u = 3 + at = is,and the gradient of = u at u = 9is8. The gradient of = ( 3 + ) at = is6. The chain rule is used to confirm this. du d = 3 and at = du d = d du = u and at u = 9 d du = 8 d d = d du du d = 6

6 Chapter Differentiation Techniques 587 Eample 7 Find the derivative of = (3+ 4) 0. Let u = then = u 0 So Eample 8 du d = 3 d d = d du du d = 0u 9 3 = 60(3 + 4) 9 d and du = 0u9 Find the gradient of the curve with equation = 6 at the point (, 4). 3 + Let u = 3 + then = 6u So du d = 6 d d = d du du d = 6u 6 = 96 (3 + ) d and du = 6u at = the gradient is 96 6 = 6. Eample 9 Differentiate = (4 3 5). Let then and u = = u du d = 5 d d = d du du d and = (u 3 ) ( 5) = ( 5) (4 3 5) 3 d du = u 3

7 588 Essential Mathematical Methods &CAS Eample 0 Use the chain rule to prove d d = nn for = n and n anegative integer. (Assume the result for n a positive integer and > 0.) Let n be a negative integer and = n. Then = as n is a positive integer. n Let u = n then = u = u Thus du d = n n ( n is a positive integer) d d = d du du d = u ( n n ) = n n ( n ) = n n ( n ) = n n and d du = u With function notation the chain rule is stated as: ( f g) () = f (g())g () where f g() = f (g()). Eample Given that f () = ( + ) 3, find f (). Now f = k g where k() = 3 and g() = + It follows that k () = 3 and g () = B the chain rule f () = k (g())g () We have in this case f () = 3(g()) which ields f () = 6( + ) Eercise B Eamples 7, 8, 9 Differentiate each of the following with respect to : a ( ) 30 b ( 5 0 ) 0 c ( 3 5 ) 4 ( d ( + + ) 4 e ( + ),, 0 f ) 3, 0

8 Chapter Differentiation Techniques 589 a Find the derivative of f : R R, f () = ( 3 + ) 4. b Find the gradient of f : R R, f () = ( 3 + ) 4 at the point (, 8). 3 a Find the gradient of the curve with equation = + 3 at the point ( ), 4. b Find the gradient of the curve with equation = ( + 3),atthe point ( ), The diagram is a sketch graph of a function with f () = + 3. = 3 a b Find the gradient of the curve at the points ( 0, 3). Find the coordinates of the points on the curve for which the gradient is , 3 5 The diagram is a sketch graph of the functions = and =. a b Find the gradient of = at the point (, ). Without further calculation state the gradient of = at ( ),. 4 3 = P(, ) Q(, ) c Find the equation of the tangent at the point (, ) of =. = = d Find the equation of the tangent at the point (, ) of =. e f Find the equations of the tangents at points P and Q and find their point of intersection. Draw sketch graphs of = and = on the same set of aes and draw in the four tangents..3 Differentiating rational powers ( p /q ) Using the chain rule in the form d du = d d d du with = u = d d d d and thus d d =, for d d d d 0 Let = n,where n Z \ {0} and > 0. Then n = and d d = nn.

9 590 Essential Mathematical Methods &CAS From the above results d d = n n = ( n n = n n ) n For = n, d d = n n ; n Z \{0} and > 0. This result ma now be etended to an rational power. p Let = q,where p, q Z \ {0}. ( ) p Write = q. Let u = q. Then = u p. The chain rule ields d d = d du du d = pu p q q ( = p = p q p q = p q p q ) p q q q q q Thus the result for an non-zero rational power has been established and, in fact, it is true for an non-zero real power. For f () = a, f () = a a, for > 0 and a R. These results have been stated for > 0, as ( 3) is not defined although ( ) 3 is. The graphs of =, = 3 and = 4 are shown. The domain of each has been taken to be R = = =

10 Chapter Differentiation Techniques 59 The figure to the right is the graph of the function f : R R, f () = 3. Note that the values shown here are From this it can be seen that the 0.4 tangent at the origin of = 3 is on the -ais. Use a calculator to investigate graphs of this tpe further. Eample = Find the derivative of each of the following with respect to : a 4 3 b 5 3 a Let = 4 3, then d d = = b Let = 5 3, then d d = = Eercise C Eample Find the derivative of each of the following with respect to : a 3 b 3, > 0 c 5 3, > 0 d 3 5 3, > 0 e 5 6, > 0 f 4, > 0 Find the derivative of each of the following with respect to : a + b 3 + c ( + ) d ( + ) 3 3 a Find the gradient of = 3 at each of the following points: ( i 8, ) ( ii ) 8, iii (, ) iv (, ) b Comment on our results. 4 Consider the graphs of = and = 3 for > 0. { } a Find : < 3. b Find the values for for which the gradient of = is greater than the gradient of = 3.

11 59 Essential Mathematical Methods &CAS 5 Differentiate each of the following with respect to : a ( 5 ) b ( 3 + ) d + e 3 ( + ) c +.4 The second derivative For f : R R, f () = + 4 +, the derived function has rule f () = 4 + 4, i.e. the derivative of is The second derivative of is4,i.e. the derivative of The function notation for the second derivative is f () = 4. In the Leibniz notation if = + 4 +, d d = and the second derivative is written d d, and d d = 4. Eample 3 For each of the following find f (). a f () = , 0 b f () = , > 0 c f () = , > 0 a f () = 9, 0 f () = , 0 b f () = 5, > 0 f () = , > 0 c f () = , > 0 f () = , > 0 If S(t) denotes the displacement of a bod at time t, then S (t)gives the velocit at time t, and S (t)gives acceleration at time t. Eercise D Eample 3 Eample 3 Find f () for each of the following: a f () = b f () = 3 + c f () = (3 + ) 4 d f () = + 3 3, > 0 e f () = ( 6 + ) 3 f f () = For each of the following find d d : a = b = 6 c = d = (6 + ) 4 e = (5 + ) 4 f =

12 Chapter Differentiation Techniques The height of a stone is (0t 4.9t ) metres at a time t seconds after it is thrown. What is the acceleration? 4 After t seconds, the -coordinate of a particle moving along the -ais is given b (t) = 4t 3t 3,where the units on the -ais are in metres. a b c Find: i the -coordinate after seconds of motion ii the velocit of the particle at the start iii the velocit of the particle after half a second iv the velocit of the particle after seconds. When is the acceleration zero? What is the average velocit during the first seconds?.5 Sketch graphs In Chapter 0 it was demonstrated how calculus ma help us sketch the graphs of polnomial functions. In this section we show how the same techniques ma be applied to non-polnomial functions. Foreample: Let f : R \{0} R, f () = +. This function is not defined at = 0. Consider the behaviour of the function as the magnitude of becomes large. This section of a spreadsheet demonstrates the behaviour of increasing. This has been done b using increasing powers of. We see that as, f (). This shows us that the line = is an oblique asmptote for the graph. It is clear also that as, f (). f() The behaviour of the function as the magnitude of approaches 0 is demonstrated below. A spreadsheet has been used and the values of considered are 0,,, 3, etc.

13 594 Essential Mathematical Methods &CAS a b f() f() Spreadsheet a shows the behaviour as approaches 0 from the right. f () becomes increasingl large. We write lim 0 + f () = Spreadsheet b shows the behaviour as approaches 0 from the left. Again f () becomes increasingl large. We write lim 0 f () =. Consider the ais intercepts. The function cannot have a -ais intercept, as it is not defined at = 0. It would cross the -ais when + = 0, i.e. when 3 =, which implies =. 3 Consider points of zero gradient (turning points). For f () = + we have f () = + and f () = 0 implies = 3 3, i.e. there is a point on the curve at = 3 at which the gradient is zero. As the graph is not continuous for R, itisadvisable to use a gradient chart for > 0 onl. f () shape of f Thus there is a minimum at the point (.6,.89) (coordinates correct to decimal places). The spreadsheet above shows how rapidl the graph of = + moves towards =. At = 4, f () = and at = 8, f () = = (.6,.89) 0 4 6

14 Chapter Differentiation Techniques 595 Eample 4 For f : R \{ } R, f () = + 3,sketch the graph. + B division, f () = Behaviour as magnitude of increases As, f (),from above. As, f (),from below. There is an oblique asmptote with equation =. Behaviour as lim f () = and lim + There is a vertical asmptote =. f () = Ais intercepts When = 0, f () = 3. There is no -ais intercept as for all R. Turning points For f () = f 4 () = ( + ) Where f () = 0 4 ( + ) = 0 which implies ( + ) = 4 i.e. + = or + = = or = 3 f () = and f ( 3) = 6 We consider two gradient charts. First for >. Net, for <. f () 0 + f () shape of f shape of f a minimum at (, ) a maimum at ( 3, 6)

15 596 Essential Mathematical Methods &CAS We are now in a position to sketch the graph. = (0, 3) ( 3, 6) 0 5 = = Eercise E The equation of a curve is = 4 +. a Find the coordinates of the turning points. b Find the equation of the tangent to the curve at the point where =. Find the -coordinates of the points on the curve = at which the gradient of the curve is 5. 3 Find the gradient of the curve = 4 4 For the curve = 5 + 4, find: at the point where the curve crosses the -ais. a the coordinates of the points of intersection with the aes b the equations of all asmptotes c the coordinates of all turning points. Use this information to sketch the curve. 5 If is positive, find the least value of For positive values of,sketch the graph of = + 4 and find the least value of. Eample 4 7 Sketch the graphs of each of the following, indicating the coordinates of the aes intercepts and the turning point, and the equations of asmptotes. a = +, 0 b =, 0 c = , 3 d = , 0 e = 5 +, 0 f = 4 +,

16 Chapter Differentiation Techniques 597 Chapter summar The general result for differentiating functions, including powers of with negative integers: For f () = n, f () = n n, n a non-zero integer. For f () =, f () = 0. We note that for n, we take the domain of f to be R \ {0}, and for n wetakethe domain of f to be R. The chain rule is often used to differentiate some more complicated functions b transforming the original function into two simpler functions: e.g. f ()istransformed to h() and g(u), which are chained together as u h g Using Leibniz notation the chain rule is stated as d d = d du du d. With function notation the chain rule is stated as ( f g) () = f (g())g () where ( f g)() = f (g()) Review The general result for an non-zero real power: For f () = a, f () = a a, for > 0 and a R. The function for the second derivative is f. In Leibniz notation it is written as d d. When using calculus as an aid to sketching graphs of polnomial and non-polnomial functions, the following should be considered: The behaviour of the function as the magnitude of becomes large The ais intercepts 3 The points of zero gradient (turning points). Multiple-choice questions If f () = 4 4 then f () equals 3 A B 4 4 C D 4 8 E p If f () = q,where p and q are integers, f () equals (p q) p A q B p q C D p (p q) q q E p q 3 For f : R \{} R,where f () = 4 + 4, f () > 0 for A R \ {} B R C < D > E > 4

17 598 Essential Mathematical Methods &CAS Review 4 A particle moves in a straight line so that its position cm from a fied point O at time t s(t 0) is given b = t 3 + 7t 4t + 6. The particle s acceleration at t = 3is A 4 cm/s B 3 cm/s C 4 cm/s D 8 cm/s E 0 cm/s 5 Let = f (g()), where g() = 3, then d d equals A 3 f ( 3 ) B 3 f ( 3 ) C f() f ( 3 ) D f () f ( 3 ) E 3 6 The graph defined b the rule f () = + has a local minimum at (a, f (a)). The value of a is A B C 5 D 5 E 7 Which of the following is not true for the curve of = f (), where f () = 5? A The gradient is defined for all real numbers. B The curve passes through the origin. C The curve passes through the points with coordinates (, ) and (, ). D For > 0 the gradient is positive. E For > 0 the gradient is decreasing. 8 Which of the following is not true for = f (), where f () = 3 4? A The maimal domain of the function is R + {0}. B f () > for all >. C The curve of = f () passes through the points with coordinates (, ). D For > 0 the gradient of the curve is positive. E For > 0 the gradient of the curve is decreasing. 9 The derivative of (5 + ) n is A n(0 + )(5 + ) n B (5 + ) n C (0 + ) n D n(5 + ) n E 0 n + n 0 The graph of the function with rule = k has gradient when =. ( + ) The value of k is A B C 4 D 4 E 4 Short-answer questions (technolog-free) Find the derivative of each of the following with respect to : a 4 b 3 c d e f g h Find the derivative of each of the following with respect to : a 3 b c d 4 3 e 3 f

18 Chapter Differentiation Techniques Differentiate each of the following with respect to : a ( + 3) b (3 + 4) 4 c (3 ) d e ( ) 3 f 3 + g ( 3 ) 3 4 Find the gradient of each of the following curves at the given point: a = ;(9, 3) b = + ;(0, ) c = ; 3 + ( 4, ) 8 Review d = 3 + ;(, 5) e = + ; (8, 3) f = ( 7 8) 3 ;(8, 0) 5 Find the coordinates of the point(s) on the curve with equation = for which the gradient is 4. 6 Find the coordinates of the point(s) on the curve with equation = for which the gradient is. Etended-response questions A solid circular clinder has radius r cm and height h cm. It has a fied volume of 400 cm 3. a Find h in terms of r. b Show that the total surface area, A cm,ofthe clinder is given b A = r r. c Find da dr. d Solve the equation da = 0 for r. dr e Find correct to 3 significant figures the minimum surface area of the clinder. f Sketch the graph of A against r. A rectangle has sides of length cm and cm and the area of the rectangle is 6 cm. a Find in terms of. b Show that the perimeter, P cm, is given b P = + 3. c Find the value of for which the value of P is a minimum and find this value of P. d Sketch the graph of P against for > 0. 3 The area of rectangle OABC is 0 cm. Let the length of OC be cm, CZ = 5cmand AX = 7 cm. X Y a Find the length of OA in terms of. A B b Find the length of OX in terms of. c Find the length of OZ in terms of. Z d Find the area, A cm,ofrectangle OXYZ in terms of. O C e Find the value of for which the area, A cm,isaminimum.

19 600 Essential Mathematical Methods &CAS Review 4 The curve with equation = + meets the -ais at A and the -ais at B. a Find the coordinates of A and B. b B using the chain rule find d d. c i Find the gradient of the curve where =. ii Find the equation of the tangent at the point where =. iii If the tangent meets the -ais at C and the -ais at D, find the distance CD. d Find the values of for which d d <. 5 An open rectangular bo of height h cm has a horizontal rectangular base with side lengths cm and cm. The volume of the bo is 36 cm 3 : a Epress h in terms of. b Show that the total surface area of the bo is given b A = c Calculate the values of and h which make the total surface area a minimum. d Sketch the graph of A against for > 0. 6 The prism shown in the diagram has a triangular cross-section. The ends of the prism shown are congruent right-angled triangles with the right angles at C and Z. AX = CZ = BY = cm, AC = XZ = 3 cm and CB = ZY = 4 cm The volume of the prism is 500 cm 3. a Epress in terms of. b Show that the total surface area, S cm,is given b S = c Find ds d. d Find the minimum value of S. A X C Z B Y