This Topic follows on from Calculus Topics C1 - C3 to give further rules and applications of differentiation.

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1 CALCULUS C Topic Overview C FURTHER DIFFERENTIATION This Topic follows on from Calcls Topics C - C to give frther rles applications of differentiation. Yo shold be familiar with Logarithms (Algebra Topic A), Eponential Fnctions (Algebra Topic A) basic Trigonometric Fnctions (Trigonometr Topic T). LESSON PLAN Lesn No. LESSON TITLE Page Eamples SAQs C. The Eponential Fnction. C. C. The Natral Logarithm Fnction 4. C. C. Trigonometric Fnctions 5. C. 4 C.4 The Chain Rle.4 C.4 5 C.5 Instantaneos Rate of Change C.5 Appendi C. Soltions to the Self-Assessment Qestions Page

2 CALCULUS C Topic Overview SUMMARY TABLE OF STANDARD DERIVATIVES FUNCTION f() DERIVATIVE f'() Power n n n Constant (C) C 0 Eponential fnction e a a e a Natral logarithm ln(a) Trigonometic fnctions sin(a) a cos(a) cos(a) tan(a) a sin(a) a sec (a) where a is a constant. Page

3 C. The Derivative of the Eponential Fnction C. THE DERIVATIVE OF THE EXPONENTIAL FUNCTION The eponential fnction e a, where a is constant, is a fnction whose rate of change is proportional to its own magnitde. [Ref: Algebra Topic A Eponentials] The rle for differentiating the eponential fnction is a a f() e f '() a e where a is a constant. Note: The special case when a, f() e f '() e EXAMPLE. Differentiate: (i) e 4 (ii) e (iii) e e Soltion: (i) Appl the rle to f() e 4 f ' () 4e 4 (ii) Use the constant mltiple rle to differentiate f() g() where g() e [Ref: C.] f() e f '() ( e ) e (iii) Appling the sm of fnctions rle to f() e e f '() e ( )e e e SAQ C. Differentiate the following fnctions with respect to the variable: (a) e (b) e (c) e t e t Page

4 C. The Derivative of the Logarithmic Fnction C. THE DERIVATIVE OF THE NATURAL LOGARITHM FUNCTION The natral logarithm fnction (log e ), sall written as ln, is the inverse of the eponential fnction. The rle for differentiating the natral logarithm fnction is f() ln(a) f ' () where a is a constant a > 0 Note that the derivative of ln(a) does not involve a. EXAMPLE. Differentiate (i) ln (ii) ln (iii) 5ln Soltion: (i) f() ln f '() (ii) Using the constant mltiple rle to differentiate f() g() where g() ln [Ref: C.] f() ln f '() (iii) Using the constant mltiple rle to differentiate f() 5g() where g() ln f() 5ln f '() 5 5 SAQ C. Differentiate the following fnctions with respect to the variable: (a) ln (b) 4 ln(t) (c) ln Page 4

5 C. The Derivative of Trigonometric Fnctions C. THE DERIVATIVE OF TRIGONOMETRIC FUNCTIONS The rles for differentiating the basic trigonometric fnctions are f() sin(a) f() cos(a) f() tan(a) f ' () a cos(a) f ' () a sin(a) f ' () a sec (a) where a is a constant [Ref: Topic T. Trigonometric Ratios] sec() cos() EXAMPLE. Differentiate (i) sin t (ii) 5cos 4t (iii) sin πt cos πt Soltion: (i) f(t) sin t f '(t) cos t (ii) Using the constant mltiple rle to differentiate f(t) 5 g(t) where g(t) cos 4t f(t) 5 cos 4t f '(t) 5( 4sin 4t) 0sin 4t (iii) Appling the sm of fnctions rle to f(t) sin πt cos πt f '(t) (π cos πt) ( π sin πt) π cos πt π sin πt SAQ C. (a) Differentiate the following fnctions with respect to the variable: (i) sin (iii) tan t (ii) cos 5 (iv) 5(sin πt cos πt) (b) An alternating voltage is given b V 4 sin 00t where t is the time in seconds. Calclate the rate of change of voltage when t 0.0 seconds. Page 5

6 C.4 The Chain Rle C.4 THE CHAIN RULE for Differentiation This rle is sed to differentiate composite fnctions sch as, cos(5 ), which can be written as h(), where g(). sin e, The Chain Rle is: where is a fnction of is a fnction of. This process is al known as the "fnction of a fnction rle". EXAMPLE.4 Using the Chain Rle, differentiate the following with respect to the variable: (i) ( 4) (ii) (iii) cos(5 ) (iv) sin (v) (ln 4) (vi) e sin t Soltion: Method : Differentiation b sbstittion (i) If ( 4) we sbstitte 4, then. Hence, b differentiating with respect to, we obtain differentiating with respect to, we obtain. Now we can appl the formla for the Chain Rle to obtain. Then ( 4) 4, ( ) ( ) ( 4) Page

7 C.4 The Chain Rle Page ( ) Then, (ii) ) 5sin(5 )(5) sin ( sin 5 Then cos, 5 ) cos(5 (iii) cos sin ()(cos ) cos Then sin, (sin ) sin (iv) (ln4) ) ( Then ln4, (ln 4) (v) sin t sin t (cos t)e )(cos t) (e e cos t Then e, sin t e (vi)

8 C.4 The Chain Rle Method : Direct application of the Chain Rle Differentiate b starting from the otside fnction working inwards, alwas mltipling b the derivative of the inner fnction. This method is normall sed since it adapts easil to deal with a chain of or more fnctions of a fnction. (i) ( 4) ( 4) ( 4) ( 4) d () ( 4) (ii) ( ) ( d ) ( ) ( ) () (iii) cos(5 ) d sin(5 ) sin(5 )(5) 5sin(5 ) ( 5 ) (iv) sin (sin ) d (sin ) (sin )(cos ) sin cos ( sin ) (v) (ln 4) d (ln 4) (ln 4) (ln 4) ( ln4) (vi) sin t e ep(sin t) sin t d e sint e (cos t) ( sint) sint (cos t) e SAQ C.4 Using the Chain Rle, differentiate the following fnctions with respect to the variable: (a) ( ) 8 (b) (c) sin(t 4) (d) cos (e) ln (5 ) (f) e t Page 8

9 C.5 Instantaneos Rate of Change C.5 INSTANTANEOUS RATE OF CHANGE The real significance of the derivative of a fnction is that it provides a measre of the instantaneos rate of change of one variable with respect to another. Since speed is a familiar measre of a rate of change, we shall investigate the idea of average instantaneos rates of change with reference to speed. If an object travels at a stea speed, we can calclate the speed sing the formla: distance travelled Speed time taken If the speed is changing, this formla will onl give the average speed over a particlar interval. The following eample shows how the instantaneos speed can be calclated. EXAMPLE.5 An object moves awa from a point O in a straight line. Its distance, s metres, from O after t seconds is given b the formla s t Find (i) the average speed between t t (ii) the instantaneos speed when t Soltion: B definition: distance travelled change in distance from O δs Speed time taken change in time δt where δs represents a change in the vale of s similarl δt represents a change in the vale of t. (i) When t second, s () metre When t seconds, s () 4 metres Hence, the average speed between t t is δs 4 δt metres per second Page 9

10 C.5 Instantaneos Rate of Change (ii) The following table shows the distances of the object from O after varios times: Time (t) Distance (s) From the table we can see that the object is accelerating awa from O its speed is continall increasing. One wa of calclating the instantaneos speed when t, is to calclate average speed between t t h then allow h to approach zero. It is convenient to se fnctional notation write f(t) t instead of s t. The average speed between t t h is therefore given b δs δ(f(t)) δt δt f( h) f() ( h) ( h) () h h h h h h h h As h approaches 0, the average speed between t t h gets closer closer to, i.e. the instantaneos speed when t is metres per second. In general, the instantaneos speed after t seconds is given b the limiting δs vale of as δt tends to zero: δt δs f(t h) lim lim δt 0 δt h 0 h f(t) It follows that we can calclate the instantaneos speed when t in the above eample sing the derivative f'(t) directl: f(t) t gives f '(t) t (sing the rle for differentiating powers) Sbstitting t gives f '() Hence, the instantaneos speed when t is metres per second. Page 0

11 C.5 Instantaneos Rate of Change EXAMPLE. In testing for diabetes, it is necessar to determine how fast the bo metabolises glcose. One test involves ingesting a stard dose of glcose, waiting hor then monitoring the concentration of glcose remaining in the blood as time passes. Sppose the amont q mg of glcose remaining in cc of blood, t hors after the dose is administered is Find the glcose levels how fast the glcose was being metabolised: (i) when monitoring began at t (ii) 4 hors after the glcose was taken. Soltion: The rate of change in glcose levels is the derivative: (i) dq q 4.5 t.5 t When t, the glcose level was the metabolic rate was dq.5().5 mg per hor The negative sign means that the glcose level was decreasing b.5 mg per hor; a positive sign wold mean that the level was increasing. (ii) For hors after the dose was taken, ie. when t 4, the glcose level was: q(4) 4.5(4) the metabolic rate was: d 4.5 t d 4.5 t 4.5 t mg mg per hor q() 4.5() dq.5( mg per hor Note: After 4 hors there was less glcose in the blood the level was dropping more slowl. mg ) 4.5 mg Page

12 C.5 Instantaneos Rate of Change SAQ C.5 The temperatre s, in degrees Centigrade, in a corridor is given b s where is the distance, in metres, from a radiator at one end. (a) Find the average rate at which the temperatre changes with distance between the points represented b 0 5. (b) Calclate the instantaneos rate of change of temperatre with respect to distance at the point represented b 5. (c) Eplain wh or answers to (a) (b) are negative. Page

13 C. APPENDIX C. SOLUTIONS TO SELF-ASSESSMENT QUESTIONS SAQ C. Eponentials (a) f () e f '() e (b) f() e f'() e e (c) f(t) e t e t f '(t) e t e t SAQ C. Natral Logarithms (a) (b) (c) f() ln f(t) 4ln(t) f() ln f'() 4 f'(t) t f'(t) SAQ C. Trigonometric Fnctions (a) (i) f() sin f '() cos (ii) f() cos 5 f '() 0 sin 5 (iii) f(t) tan t f '(t) sec t (iv) f(t) 5(sin πt cos πt) 5sin πt 5cos πt f '(t) 0πcosπt 0sin πt 0π(cosπt sin πt) (b) f(t) 4 sin 00t f '(t) 400 cos 00t When t 0.0 seconds, f '(0.0) 400 cos(00 0.0) 400 cos(). volts per second SAQ C.4 The Chain Rle (a) then ( ), ( 8 )( ) or ( ) 8 8( ) ( ) () ( ) Page

14 C. APPENDIX Page 4 ( ) then, (b) cos sin sin ) )( ( sin then cos, (cos ) cos (d) 5 5 ) (5 5 then ln, 5 ) ln(5 (e) ( ) ) ) ( or ( 4) cos(t )() (cos cos then sin 4, t 4) sin(t (c) 4) cos(t 4)() cos(t 4) sin(t or cos sin sin ) )( (cos (cos ) cos or 5 5 ) (5 5 ) ln(5 or

15 C. APPENDIX (f) then t e t, (e )() e t e e or t e t (e )() t e SAQ C.5 Instantaneos Rate of Change (a) When When 400 s 0 0, 400 s 0 0 5, s Average rate at which the temperatre changes with distance between these two points is: Change in temperatre Change in distance (b) The instantaneos rate of change is given b the derivative s ( 400 ) 0 ds ( ) 0 ds When 5, ( 0) (c) The answers to (a) (b) above are both negative becase the temperatre s, is decreasing as the distance, from the radiator is increasing. If the graph of s against were to be drawn, then (at least for positive vales of ) the shape of the graph wold be a falling crve. Hence the gradient of the crve at an point wold be negative. Page 5

Differentiation of Exponential Functions

Differentiation of Exponential Functions Differentiation of Eponential Fnctions The net derivative rles that o will learn involve eponential fnctions. An eponential fnction is a fnction in the form of a constant raised to a variable power. The