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### lecture (8)derivatives.pptxdddddddddddddff

• 2. Differentiation Today’s Objective: When you have completed this lecture you will be able to: • Differentiate by using a list of standard derivatives • Apply the chain rule • Apply the product and quotient rules • Perform logarithmic differentiation • Differentiate parametric equations 2
• 3. Here is a revision list of the standard derivatives which you have no doubt used many times before. Memorize those with which you are less familiar. 3 Standard derivatives No. 1 2 3 4 5 6 7 x sin  No. 8 9 10 11 12 13 14 ) (x f y  dx dy n x 1  n nx x e x e x a a ax ln . x ln x 1 x a log a x ln . 1 x sin x cos ) (x f y  dx dy x 2 sec x cot x ec2 cos  x sec x x tan . sec ecx cos x ecx cot . cos  x sinh x cosh x cosh x sinh x tan x cos x sin  kx e kx ke
• 4. 4 Example 1: write down the derivatives for the following. Solution : x x x x e x x 2 . 6 tan . 5 ln . 4 . 3 sin . 2 . 1 3 5 x e x x x x cos . 11 . 10 log . 9 cosh . 8 sec . 7 10 x a x a ecx x . 16 cot . 15 . 14 cos . 13 sinh . 12 3 2 4 . 20 . 19 log . 18 . 17 x a e x x x 2 ln . 2 . 6 sec . 5 1 . 4 3 . 3 cos . 2 5 . 1 2 3 4 x x x x e x x x e x x x x x sin . 11 . 10 10 ln . 1 . 9 sinh . 8 tan . sec . 7  x ec anx ecx x 2 cos . 15 0 . 14 cot . cos . 13 cosh . 12   2 2 1 5 . 20 2 1 . 19 ln 1 . 18 4 . 17 ln . . 16 x x e x a x x a a  
• 5. 5 Function of a function Is a function of since the value of depends on the value of angle . Similarly is a function of the angle since the value of the sine depends on the value of this angle. i.e. Is a function of But it itself is a function of , since its value depends on . For example : the derivative of the function x sin x x sin x ) 5 2 sin(  x ) 5 2 (  x ) 5 2 sin(  x ) 5 2 (  x ) 5 2 (  x x x 5 ) 3 4 (   x y    4 ) 3 4 ( 5 x dx dy ) 3 4 (  x 4 ) 3 4 ( 5 4     x dx dy 4 ) 3 4 ( 20   x
• 6. 6 Example 2 : write down the derivatives for the following. Solution : ) cos 4 3 ln( . 5 ) cos( . 4 2 sin . 3 . 2 ) 5 4 ( . 1 2 3 6 x y x y x y e y x y x         x y x y x y e y x y x 3 cos ln . 10 ) 3 ( cos . 9 sin . 8 . 7 ) 1 2 ( log . 6 3 2 2 sin 10       x x y x x y x dx dy e dx dy x dx dy x cos 4 3 sin 4 . 5 ) sin( 2 . 4 2 cos 2 . 3 . 2 ) 5 4 ( 24 . 1 2 3 5             x x x y x x y x x y e x y x y x 3 tan 3 3 cos 3 sin 3 . 10 ) 3 ( cos ) 3 sin( 9 . 9 sin cos 2 . 8 . 2 cos 2 . 7 10 ln ) 1 2 ( 2 . 6 2 2 sin               
• 7. 7 Differentiate functions which are products or quotients of two of the functions dx du v dx dv u dx dy   1. Products : if y=uv where u and v are functions, of x, then you already know that: 2 v dx dv u dx du v dx dy   2. Quotients : if y=u/v where u and v are functions, of x, then you already know that:
• 8. 8 Example 3: write down the derivatives for the following. Solution : 2 2 2 3 5 2 2 cos . 8 ln . 7 1 3 sin . 6 sinh ln . 5 5 sin . 4 2 cos . 3 ) 1 3 ( . 2 tan . 1 x x y e x y x x y x x y x x y x x y x e y x x y x x           ) 1 3 ( 5 3 . 2 tan 2 sec . 1 5 5 2 2      x e e dx dy x x x x dx dy x x ) 5 15 ( 3 5 5    x e e x x ) 8 15 ( ) 5 15 3 ( 5 5      x e x e x x x x x dx dy 2 cos 2 sin 2 . 3    x x x x dx dy 5 sin 3 5 cos 5 . 4 2 3   ) 5 sin 3 5 cos 5 ( 2 x x x x   x x x x x dx dy sinh 2 sinh cosh . 5 2   ) sinh 2 coth ( x x x x   2 ) 1 ( 3 sin 3 cos ) 1 ( 3 . 6     x x x x dx dy x x x x e x x e x e x e dx dy 2 4 2 2 ln 2 1 ln 2 . 7     4 2 2 cos 2 2 sin 2 . 8 x x x x x dx dy    3 ) 2 cos 2 sin ( 2 x x x x   
• 9. 9 Logarithmic differentiation. The rules of differentiating a product or a quotient that we are revised are used when there are just two-factor functions, i.e. uv or u/v. where there are more than two functions in any arrangement top or bottom, the derivative is best found by what is known as logarithmic differentiation. For example; if where u, v, and w -and also y- are all functions of x. First take logs of base e w uv y  w v u y ln ln ln ln    dx dw w dx dv v dx du u dx dy y 1 1 1 1             dx dw w dx dv v dx du u y dx dy 1 1 1
• 10. 10 Example 4: find dy/dx for the following. Solution : x x e y x e x y x x x y x x 2 cosh . 3 tan . 2 2 cos sin . 1 3 4 3 4 2                                   x x x x x x dx dy x x x x x x x x x dx dy dx dw w dx dv v dx du u y dx dy x w x v x u let 2 tan 2 cot 2 2 cos sin 2 cos ) sin 2 ( sin cos 2 2 cos sin 1 1 1 2 cos , sin , . 1 2 2 2 2                               x x x x e x dx dy x x e e x x x e x dx dy dx dw w dx dv v dx du u y dx dy x w e v x u let x x x x x tan sec 3 4 tan tan sec 3 4 tan 1 1 1 tan , , . 2 2 3 4 2 3 3 4 3 3 4 3 4                               x x x x e x x x x e e x x e dx dy dx dw w dx dv v dx du u y dx dy x w x v e u let x x x x x 2 tanh 2 3 4 2 cos 2 cosh 2 sinh 2 3 4 2 cos 1 1 1 2 cosh , , . 3 3 4 3 2 4 4 3 4 3 4
• 11. 11 Parametric equations In same cases, it is convenient to represent a function by expressing x and y separately in terms of a third independent variable. E.g. y= cos2t, x=sint. The third variable is called parameter, and the two expression of x and y are called parametric equation. Example 5: find and for the following. Solution : 1. t t y t t x x y t x t y            1 2 3 , 1 3 2 . 3 cos , sin sin 3 . 2 sin , 2 cos . 1 3 3    dx dy 2 2 dx y d t dt dx t dt dy cos 2 sin 2    dx dt dt dy dt dt dx dy dx dy . .   t t dx dy cos 1 . 2 sin 2  
• 13. 13          sin cos 3 ) sin 1 ( cos 3 sin cos 3 cos sin 3 cos 3 2 2 2 2       dx dy      cot sin cos 3 cos cos 3 2 2           dx dt dt d dx y d dt dt dx d dx y d dx d dx y d . cot . cot cot 2 2 2 2 2 2                  sin cos 3 cos sin cos 3 1 ). cos ( 2 2 2 2 2 2 ec ec dx y d       Solution : 3. 2 2 2 2 2 2 ) 1 ( 5 ) 1 ( 3 2 3 3 ) 1 ( ) 3 2 ( ) 1 ( 3 ) 1 ( 1 ) 1 ( 2 3 2 2 ) 1 ( ) 2 3 ( ) 1 ( 2 t t t t t t t dt dx t t t t t t t dt dy                               dx dt dt dy dx dy . 5 1 5 ) 1 ( . ) 1 ( 1 2 2       t t dx dy 0 5 1 2 2         dx d dx y d
• 15. Exponential and logarithmic functions • Today’s Objective: When you have completed this lecture you will be able to: • Solve indicial and logarithmic equations. • Recognize that the exponential function and the natural logarithmic function are mutual inverses • Construct the hyperbolic functions from the odd and even parts of the exponential function. • Application of logarithms and exponential function. 15
• 16. Introduction to logarithms If a number y can be written in the form , then the index x is called the ‘logarithm of y to the base of a’, Logarithms having a base of 10 are called common logarithms and is usually abbreviated to . 16 10 log x a y  y x then a y if a x log   lg Laws of logarithms There are three laws of logarithms, which apply to any base: B A A n A B A B A B A n log log ) log( . 3 log ) log( . 2 log log ) log( . 1      
• 17. An indicial equation is an equation where the variable appears as an index and the solution of such an equation requires the application of logarithms. 17 Indicial equations Example 1: Find the value of X, give that 4 . 35 122  x Solution : 7177 . 0 0792 . 1 2 5490 . 1 5490 . 1 0792 . 1 ) 2 ( 4 . 35 log 12 log ) 2 ( 4 . 35 log 12 log 2         x x x x A n An log ) log( 
• 18. 18 Example 2: Find the value of X, give that Solution : 694 . 6 3913 . 0 6192 . 2 2042 . 1 4150 . 1 ) 4150 . 1 8063 . 1 ( 4150 . 1 4150 . 1 2042 . 1 8063 . 1 4150 . 1 ) 1 ( 6021 . 0 ) 2 3 ( 26 log ) 1 ( 4 log ) 2 3 ( 26 log 4 log 1 2 3                   x x x x x x x x x x log(An )= nlogA 1 2 3 26 4    x x
• 19. 19 Example 3: Find the value of X, give that Solution : 4853 . 2 5164 . 0 2834 . 1 2834 . 1 ) 0436 . 2 5600 . 2 ( 0436 . 2 2834 . 1 5600 . 2 0436 . 2 9138 . 0 8276 . 1 19726 . 2 7324 . 0 0436 . 2 9138 . 0 ) 1 2 ( 7324 . 0 ) 3 ( ) 6812 . 0 ( 3 2 . 8 log ) 1 2 ( 4 . 5 log ) 3 ( 8 . 4 log 3 2 . 8 log 4 . 5 log 8 . 4 log ) 2 . 8 4 . 5 log( 1 2 3 3 1 2 3                               x x x x x x x x x x x x x x x x x x x B A B A log log ) log(    x x x 3 1 2 3 8 . 4 2 . 8 4 . 5    
• 20. 20 Example 4: Find the value of X, give that Solution : 5006 . 1 7676 . 2 1533 . 4 1533 . 4 7676 . 2 4683 . 2 6123 . 1 7766 . 5 1553 . 1 8450 . 0 294 log 4 . 6 log 2 3 . 14 log ) 5 ( 7 log 294 log } 4 . 6 ) 3 . 14 ( 7 log{ 2 5                   x x x x x x x x 294 4 . 6 ) 3 . 14 ( 7 2 5    x x Example 5: Solve the equation Solution : 2 1 4 2 2 2 4 0 ) 4 ( 2 0 ) 2 ( 0 ) 4 )( 2 ( 0 8 6 2 2                         x x z or z z z or z z z z z z z let x x x 0 8 2 6 22     x x
• 21. Graphs of logarithmic functions 21 A graph of is shown in bellow x y 10 log  0 1 log  a   0 loga 1 log  a a In general, with a logarithm to any base a, it is noted that:
• 22. Exponential functions The exponential function is expressed by the equation: or where e is the exponential number 2.7182818 _ _ _ . The graph of this function lies entirely above the x-axis as does the graph of its reciprocal , as can be seen in the diagram: 22 ) exp(x y  x e y  x e y  
• 23. The value of can be found to any level of precision desired from the series expansion: In practice a calculator is used. Logarithms with base exponential called Napierian logarithms which is written as: So, 23 x e x x e ln log          ! 5 ! 4 ! 3 ! 2 1 5 4 3 2 x x x x x ex x e x e x x e    ln log . .......... ln 3   x e
• 24. 24 Example 6: Solve for X, given that Solution : ) 1 ( 60 20 2 x e   810 . 0 405 . 0 2 405 . 0 2 2 3 ln ln 2 3 3 2 6 2 1 ) 1 ( 60 20 2 2 2 2 2                    x x x e e e e e x x x x x Example 7: Solve for X, given that Solution : 0 2 3 2    x x e e 0 ) 2 )( 1 ( 0 2 3 2          y y y y e y let x 0 1 ln 1 ln ln 1 1 0 ) 1 (         x x e e y y x x 693 . 0 2 ln 2 ln ln 2 2 0 ) 2 (         x x e e y y x x
• 25. 25 Example 8: Solution : Ans. 0.5545s The current amperes flowing in a capacitor at time seconds is given by where and Determine : (a) the time for the current to reach 6.0 A. (b) Sketch the graph of current against time. ) 1 ( 8 CR t e i    i t    3 10 25 R f C 6 10 16    t
• 26. 26 Odd and even parts Not every function is either even or odd but many can be written as the sum of an even part and an odd part. If, given f(x) where f( -x) is also defined then: is even and is odd Furthermore is called the even part of f(x) and is called the odd part of f(x). For example, if then So that the even and odd parts of are: 2 ) ( ) ( ) ( x f x f x fe    2 ) ( ) ( ) ( x f x f x fo    ) (x fe ) (x fo 1 2 3 ) ( 2    x x x f 1 2 3 1 ) ( 2 ) ( 3 ) ( 2 2          x x x x x f ) (x f x x x x x x f x x x x x x f o e 2 2 ) 1 2 3 ( ) 1 2 3 ( ) ( 1 3 2 ) 1 2 3 ( ) 1 2 3 ( ) ( 2 2 2 2 2                 So, the even and odd parts of the odd part of are……………… 4 3 2 ) ( 2 3     x x x x f
• 27. 27 Odd and even parts of the exponential function The exponential function is neither odd nor even but it can be written as a sum of an odd part and an even part. That is, and . These two functions are known as the hyperbolic cosine and the hyperbolic sine respectively: and Using these two functions the hyperbolic tangent can also be defined: The logarithmic function is neither odd nor even and indeed does not possess even and odd parts because is not defined. 2 ) exp( ) exp( ) ( exp x x x e    2 ) exp( ) exp( ) ( exp x x x o    2 cosh x x e e x    2 sinh x x e e x    x x x x e e e e x      tanh x y a log  ) ( log x a 

### Editor's Notes

1. cos2x=cos↑2x-sin↑2x sin2x=2sinxcosx
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