Exponential and Logarithm functions : Exponential functions : If a is a positive number and x is any number , we define the exponential function as : y = ax with domain : -∞ < x < ∞ Range : y > 0 The properties of the exponential functions are : 1. If a > 0 ↔ ax > 0 . 2. ax . ay = ax + y . 3. ax / ay = ax - y . 4. ( ax )y = ax.y . 5. ( a . b )x = ax . bx . x 6. a y a x ( a ) x . 7. a-x = 1 / ax and ax = 1 / a-x . 8. ax = ay ↔ x = y . 9. a0 = 1 , a∞ = ∞ , a-∞ = 0 , where a > 1 . a∞ = 0 , a-∞ = ∞ , where a < 1 . The graph of the exponential function y = ax is : y y Logarithm function : If a is any positive number other than 1 , then the logarithm of x to the base a denoted by : y = logax where x > 0 At a = e = 2.7182828… , we get the natural logarithm and denoted by : y = ln x Let x , y > 0 then the properties of logarithm functions are : 1. y = ax ↔ x = logay and y = ex ↔ x = ln y . 2. logex = ln x . 3. logax = ln x / ln a . ١ 4. 5. 6. 7. 8. 9. ln (x.y) = ln x + ln y . ln ( x / y ) = ln x – ln y . ln xn = n. ln x . ln e = logaa = 1 and ln 1 = loga1 = 0 . ax = ex. ln a . eln x = x . The graph of the function y = ln x is : y2 1.5 1 0.5 0 -0.5 0 1 2 e X 3 4 5 -1 -1.5 -2 Application of exponential and logarithm functions : We take Newton's law of cooling : T – TS = ( T0 – TS ) et k where T is the temperature of the object at time t . TS is the surrounding temperature . T0 is the initial temperature of the object . k is a constant . EX-1- The temperature of an ingot of metal is 80 oC and the room temperature is 20 oC . After twenty minutes, it was 70 oC . a) What is the temperature will the metal be after 30 minutes? b) What is the temperature will the metal be after two hours? c) When will the metal be 30 oC? Sol. : T T S ( T0 T S )e tk 50 60 e 20 k k ln 5 ln 6 0 .0091 20 a) T 20 60 e 30 ( 0.0091 ) 60 * 0.761 45 .6 o C T 65.6 o C b) T TS 60 e 120 ( 0.0091 ) 60 * 0.335 20.1 o C T 40 .1 o C c) 10 60 e 0.0091 t 0.0091 t ln 6 t 3.3 hrs . ٢ Exponential functions : If u is any differentiable function of x , then : d u du a a u .ln a . dx dx EX-2 –Find d u du e eu . dx dx and dy for the following functions : dx a ) y 2 3x b) y 2 x .3 x c) y ( 2 x ) 2 d ) y x.2 x 5x e) y e (x e ) f ) ye 2 1 5x 2 Sol.dy 2 3 x * 3 ln 2 dx dy b ) y 2 x .3 x y 6 x 6 x . ln 6 dx dy c ) y (2 x )2 y 2 2 x 2 2 x ln 2.2 2 2 x 1 ln 2 dx 2 2 2 2 dy d ) y x.2 x x.2 x ln2.2x 2 x 2 x (2x 2 ln2 1) dx 5x dy (x e 5x ) e) ye e (x e ) ( 1 5 e 5 x ) dx 1 1 1 2 2 2 2 1 2 dy f ) y e (1 5x ) e (1 5x ) ( 1 5 x 2 ) 2 .10 x e 1 5 x dx 2 a ) y 2 3x 5x 1 5x2 Logarithm functions : If u is any differentiable function of x , then : 8) d 1 du loga u . dx u .ln a dx EX-3 – Find d 1 du ln u . dx u dx and dy for the following functions : dx a ) y log10 e x b ) y log 5 ( x 1 )2 c ) y log 2 ( 3 x 2 1 )3 d ) y ln(x 2 2 )2 e ) y ln(xy) 1 (2x 3 4 ) 3 .( 2 x 2 3 ) 2 f) y ( 7 x 3 4 x 3 )2 2 3 3 5 Sol. – dy log 10 e ln e 1 dx ln 10 ln 10 dy 2 b ) y log 5 ( x 1 )2 2 log 5 ( x 1 ) dx ( x 1 ) ln 5 dy 3 18 x c ) y 3 log 2 ( 3 x 2 1 ) . 6x dx 3 x 2 1 ln 2 ( 3 x 2 1 ) ln 2 2 2 dy 48 x ln( x 2 2 ) 2 2 d) 3 2 ln( x 2 ) .2 x dx x2 2 x2 2 a ) y log 10 e x y x log 10 e dy 1 1 dy dy y . 0 dx x y dx dx x( y 1 ) 2 5 f ) lny ln( 2 x 3 4 ) ln( 2 x 2 3 ) 2 ln( 7 x 3 4 x 3 ) 3 2 2 1 dy 2 6 x 5 4x 21 x 2 4 . . 3 . 2 2. 3 y dx 3 2 x 4 2 2 x 3 7 x 4x 3 2 dy 2x 5x 21 x 2 4 2 y 3 2 3 dx 2x 4 2x 3 7 x 4x 3 e ) y lnx lny 1 4 Hyperbolic functions : Hyperbolic functions are used to describe the motions of waves in elastic solids ; the shapes of electric power lines ; temperature distributions in metal fins that cool pipes …etc. The hyperbolic sine (Sinh) and hyperbolic cosine (Cosh) are defined by the following equations : 4. 1 u 1 ( e e u ) and Coshu ( e u e u ) 2 2 Sinhu e u e u Coshu e u e u tanh u and Cothu Coshu e u e u Sinhu e u e u 1 2 1 2 Sechu u and Cschu u u Coshu e e Sinhu e e u 2 2 Cosh u Sinh u 1 5. tanh 2 u Sech 2 u 1 1. Sinhu 2. 3. and Coth 2 u Csch 2 u 1 Coshu Sinhu e u and Coshu Sinhu e u Cosh( u ) Coshu and Sinh( u ) Sinhu Cosh0 1 and Sinh0 0 Sinh( x y ) Sinhx .Coshy Coshx .Sinhy Cosh( x y ) Coshx .Coshy Sinhx .Sinhy Sinh 2 x 2.Sinhx .Coshx Cosh 2 x Cosh 2 x Sinh 2 x Cosh 2 x 1 Cosh 2 x 1 13. Cosh 2 x and Sinh 2 x 2 2 6. 7. 8. 9. 10. 11 . 12. y=Sinhx y=Cschx y=Cschx 0 5 y y=Coshx 1 1 1 1 1 y=Sechx 0 0 0 0 y=Cothx 1 y=tanhx 0 y=Cothx y Sinhx y Coshx y tanh x y Cothx y Sechx y Cschx -1 D x : x and D x : x and D x : x and D x : x 0 and D x : x and D x : x 0 and R y : y Ry : y 1 R y : 1 y 1 R y : y 1 or y 1 Ry : 0 y 1 R y : y 0 EX-1 - Let tanh u = - 7 / 25 , determine the values of the remaining five hyperbolic functions . Sol.- 6 Cothu 1 25 tanh u 7 tanh 2 u Sech 2 u 1 1 25 Sechu 24 Sinhu tanh u Coshu 49 24 Sech 2 u 1 Sechu 625 25 Coshu Cschu 7 Sinhu 7 Sinhu 25 25 24 24 1 24 Sinhu 7 EX-2 - Rewrite the following expressions in terms of exponentials . Write the final result as simply as you can : a ) 2Cosh(ln x ) b ) tanh(ln x ) c ) Cosh5 x Sinh5 x d ) ( Sinhx Coshx )4 Sol.e ln x e ln x 1 a ) 2Cosh(ln x ) 2. x 2 x 1 x 2 e ln x e ln x x x 1 b ) tanh(ln x ) ln x 1 x2 1 e e ln x x x 5 x 5x e e e 5 x e 5 x c ) Cosh5 x Sinh5 x e5x 2 2 4 x x x e e e ex 4 e 4 x d ) ( Sinhx Coshx ) 2 2 EX-3 - Solve the equation for x : Cosh x = Sinh x + 1 / 2 . 1 1 Sol. - Coshx Sinhx e x x ln 1 ln 2 x ln 2 2 2 EX-4 – Verify the following identity : a) Sinh(u+v)=Sinh u. Cosh v + Cosh u.Sinh v b) then verify Sinh(u-v)=Sinh u. Cosh v - Cosh u.Sinh v Sol.7 a ) R .H .S . Sinhu .Coshv Coshu .Sinhv e u e u e v e v e u e u e v e v . 2 2 2 2 u v ( u v ) e e Sinh( u v ) L .H .S . 2 b ) L .H .S . Sinh( u ( v )) Sinhu .Cosh( v ) Coshu .Sinh( v ) Sinhu .Coshv Coshu .Sinhv R .H .S . EX-5 – Verify the following identities : 1 a) Sinhu .Coshv Sinh( u v ) Sinh( u v ) 2 1 b ) Coshu .Coshv Cosh( u v ) Cosh( u v ) 2 3 c ) Sinh 3 u Sinh u 3Cosh 2 u .Sinhu 3 Sinhu 4 Sinh 3 u d ) Sinh 2 u Sinh 2 v Cosh 2 u Cosh 2 v Sol. – 1 a ) R .H .S . Sinh( u v ) Sinh( u v ) 2 1 Sinhu .Coshv Coshu .Sinhv Sinhu .Coshv Coshu .Sinhv 2 Sinhu .Coshv L .H .S . 1 b ) R .H .S . Cosh( u v ) Cosh( u v ) 2 1 Coshu .Coshv Sinhu .Sinhv Coshu .Coshv Sinhu .Sinhv 2 Coshu .Coshv L .H .S . c ) L .H .S . Sinh( 2 u u ) Sinh 2 u .Coshu Cosh2 u .Sinhu 2 Sinhu .Coshu .Coshu ( Cosh 2 u Sinh 2 u ).Sinhu 3 Sinhu .Cosh 2 u Sinh 3 u R .H .S .( I ) 3 Sinhu .( 1 Sinh 2 u ) Sinh 3 u 3 Sinhu 4 Sinh 3 u R .H .S .( II ) d ) L .H .S . Sinh 2 u Sinh 2 v Cosh 2 u 1 ( Cosh 2 v 1 ) Cosh 2 u Cosh 2 v R .H .S . Inverse hyperbolic functions : All hyperbolic functions have inverses that are useful in integration and interesting as differentiable functions in their own right . 8 0 y Sinh 1 x 1 y Cosh 1 x D x : x R y : y -١ -1 ١ y tanh1 x Dx : 1 x 1 R y : y D x : x 1 R y : y 0 ١ y Coth1 x Dx : x 1 or x 1 Ry : y 0 ١ y Sech 1 x y Csch1 x Dx : x 0 R y : y 0 Dx : 0 x 1 R y : y 0 9 Some useful identities : Sinh 1 x ln( x x 2 1 ) Cosh 1 x ln( x x 2 1 ) 1 1 x 3. tanh 1 x . ln 2 1 x 1 x 1 1 1 4. Coth 1 x . ln tanh 2 x 1 x 2 1 1 x Cosh 1 1 5. Sech1 x ln x x 1. 2. 1 6. Csch x ln x 1 x 2 1 1 Sinh 1 x x EX-6 - Derive the formula : Sinh 1 x ln( x x 2 1 ) Sol.e y e y e2y 1 Let y Sinh x x Sinhy x 2 2e y e 2 y 2 x .e y 1 0 2x 4 x2 4 y e e y x x2 1 2 2 either y ln( x x 1 ) neglected sin ce x x 2 1 0 1 or y ln( x x2 1 ) 10