Differentiation. The main problem of differential calculus deals with finding the slope of the tangent line at a point on a curve.

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1 Dierentiation The main problem o dierential calculus deals with inding the slope o the tangent line at a point on a curve. deinition() : The slope o a curve at a point p is the slope, i it eists, o the tangent line at p. deinition () : The derivative o a unction is the unction denoted (read prime) and deined by : () = = ( ) ( ) ( ) ( ) provided that this limits eist. i (a) can be ound ( while perhaps not all () can be ound ) is said to be dierentiable at a, and (a) is called the derivative o with respect to at a. the process o inding the derivative is called dierentiation. eample : using the deinition to ind the derivative : i () =, ind the derivative o solution : applying the deinition o a derivatives gives : () = ( ) ( )

2 = ( )( ) = ( ) = =. Observe that, in taking the limits, we treated as a constant, because it was h not. besides the notation F (), there are several ways to denote the derivative y = () at., pronounced dee y by dee. (()), dee (), dee. Note : Because the derivative gives the slope o the tangent line, (a) is the slope o the tangent to the graph o y = () at (a, (a)). : Find the slope o the curve y =+ at the point where = 6? Solution : The slope o the curve is the slope o the tangent line. letting y = ()= + ( ) ( ) = ( ) ( ) = = =.

3 So the slope when =6, or in act at any point is, because it is constant. Deinition : I is dierentiable at a, then is continuous at a. For eample :. Let () =, the derivative, is deined or all values o, so () = must be continuous or all values o.. The unction (p)= is not continuous at p =0 because is Note: not deined there,thus, the derivative does not eist at p=0. Continuity does not imply dierentiability: The unction y =() = is continuous at =0,it continuous at =0, but there is no tangent line at =0. thus the derivative does not eist there.this shows that continuity does not imply dierentiability.

4 Problems : Use the deinition o the derivative to ind each o the ollowing :. F (), i ()=.. F () i () = 4-.. F () i y = F() i () =. 4

5 Rules or dierentiation : Dierentiating a unction by direct use o the deinition o derivative can be tedious, we have to simpliy the operation in order to skip time. So we have to know the derivative o a ew basic unctions and ways to assemble derivatives o constructed unctions rom the derivatives o their components. Basic rule 0: derivative o a constant : I c is a constant, then (c) = 0 That is, the derivative o a constant unction is a zero. For eample :. ( ) = 0. ( ) = 0 Basic rule :derivative o n : I a is any real number, then For eample :. ( 5 ) = 5 5- = 5 4. ( n )= n n-. 5

6 Combining rule : constant actor rule : I is a dierentiable unction and c is a constant, then c * () is dierentiable, and : ( ( )) ( ) that is, the derivative o a constant times a unction is the constant times the derivative o the unction. or eample : y = () =5 () = 5 combining rule : Sum or dierence rule : i and g are dierentiable unctions, then +g and g are dierentiable and : And ( ( ) ( )) ( ) ( ) ( ( ) ( )) ( ) ( ) That is, the derivative o the sum (dierence ) o two unctions is the sum (dierence )o their derivative. 6

7 For eample : Given Y = () = 5 + Find ( ) Solution : ( ) : Find ( ) Solution : Y = () = () = Rule number : Product rule: i y = () * g() then : = () * + * g() 7

8 note that ; in case o dierence just the sign o the sum will be changed. Applying the product rule : i y =( +) ( - +5). ind. solution : = ( +) ( - +5) + ( +) ( - +5) = ( +)(5- -4 ) + ( - +5) ( - ) = Hint :In case o dierentiating a product o three actors,we combine two o them into one and continue to dierentiate with the previous rule. : I y = (+)(+)(+4) Solution : First we combine two actors through multiplication Take: (+)(+) =( ) Then y will be = ( +5+6)(+4) = ( +5 +6) (+4) + ( +5+6) (+4) = ( +5) (+4) + ( ) * =( ) + ( +5 +6) =

9 Now ater you know how to use the multiply rule, you can use it to ind the slope o any rule that includes a chance o multiplying unctions. Combining rule 4 : the Quotient rule I and g and dierentiable unctions and g() 0, then the quotient g is also dierentiable, and : ( ( ) ) = ( ) ( ) ( ) ( ) ( ) You must understand that the dominator should not zero. That is, the derivative o the quotient o two unctions is the denominator times the derivative o the numerator minus the numerator times the derivative o the denominator, all divided by the square o the denominator. : Solution : I () =, ind () () = ( ) ( ) ( ) = ( ) ( ) ( ) 9

10 Dierentiating quotients without using the quotient rule a. () = by rewriting we have () = so () = = b. () = by rewriting we have () = = ( ) = ( ) ( ) () =. ) The chain Rule : Our net rule, the chain rule, is ultimately the most important rule or inding derivatives. It involves a situation in which y is a unction o the variable u, but u is a unction o, and we want to ind the derivative o y with respect to. 0

11 For eample, the equation s : Y = u and u = + Deine y as a unction o u and u a unction o.i we substitute + or u in the irst equation, we can consider y to be a unction o ; Y =( + ) And in order to ind, we have to epand( + ) Y = () = Then, = Combining rule 5 : the chain rule : I y is a dierentiable unction o u and u is a dierentiable unction o, then y is a dierentiable unction o. Like in the previous eample : = u = (+) And = = ( + ) * = 4 ( + ) = (4) : Using the chain rule I y = u u - and u = + 4, ind

12 Solution: ( ) ( ) =(4u ) * () Since u = +4 Then =( 4 ( +4) - ) () =( ) () = The Power rule : (u a ) = a u a- Where it is understood that u is a dierentiable unction o and a is a real number. (4) : Using the power rule, ind Solution : I y = ( ) Since y = (4 + - ) Through comparing with the power rule a =

13 And u =4 + = *(4 + - ) - ( ) *(4 + - ) - (8 + ) = ( ) Y =( +6) 00 We think o the unction as a composition. let Y = (u) and u = g() = + 6 ( ) ( ) As you can see, we have just used the chain rule to dierentiate y =( + 6) 00, which is a power o unction o, not simply a power o. - y = u 8, u = - dy d dy d dy du 8 u 7. du d 7 6u dy d 7 6

14 4 Eercise () : - In the same way try to ind, i y = u +4, And u = 5 - Find the derivative or the ollowing unctions L Z W W Y ind the partial derivative dy/d - dy/d 4... Y iii Y ii Y i

15 Dierentiation o Logarathmatic unction: I y= L n () dy d / y= In () = ln / dy d dy d y In I y = Ln ( + + ) Find Solution ( ) ( ) I z = Ln (t + 5t + ) Find 5

16 IF y = Ln ( ) 5 solution Use the rules o the logarathmatic unction y = 5Ln( ) ( ) I [ ] ind / (). solution Use the rules o the logarathmatic unction y= Ln( +) Ln( +) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) Dierentiation o the Eponential unction: 6

17 With base e I y = e () dy / d Then e e Y= dy d / e e Y= e dy d dy d / e 9 4 e 7

18 Higher Order Derivatives How to Find High-Order Derivatives Finding a second, third, ourth, or higher derivative is incredibly simple. The second derivative o a unction is just the derivative o its irst derivative. The third derivative is the derivative o the second derivative, the ourth derivative is the derivative o the third, and so on. Let s start this section with the ollowing unction. By this point we should be able to dierentiate this unction without any problems. Doing this we get, Now, this is a unction and so it can be dierentiated. Here is the notation that we ll use or that, as well as the derivative. 8

19 This is called the second derivative and now called the irst derivative. is Again, this is a unction so we can dierentiate it again. This will be called the third derivative. Here is that derivative as well as the notation or the third derivative. Continuing, we can dierentiate again. This is called, oddly enough, the ourth derivative. We re also going to be changing notation at this point. We can keep adding on primes, but that will get cumbersome ater a while, For eample, here s a unction and its irst, second, third, and subsequent derivatives. In this eample, all the derivatives are obtained by the power rule: 9

20 A irst derivative tells you how ast a unction is changing how ast it s going up or down that s its slope. A second derivative tells you how ast the irst derivative is changing or, in other words, how ast the slope is changing. A third derivative tells you how ast the second derivative is changing, which tells you how ast the rate o change o the slope is changing. I you re getting a bit lost here, don t worry about it. It gets increasingly diicult to get a handle on what higher derivatives tell you as you go past the second derivative, because you start getting into a rate o change o a rate o change o a rate o change, and so on. 0

21 Eercise Find the derivative o the unction (a) - (b) y = (c) y = e (d) y = e t+4 (e) y = Ln(t+) () y = Ln 8t (g) y = Ln rom the unction ind : (a) (b) y = Ln( + ) (c) - () y = ( + )( - ) Using the chain rule ind () z = (y + 5y + ) 8 () z = ( y + 4) /5 Find the total derivative orm the unctions: - y = (,w) = w - y = g(w) = w + w ( ) 4- y = (,z) = + z + z

22 Integration I we have y= (), then / ( ) = d d This what we call dierentiation. While Integration is to return back to the original unction () d d - Indeinite Integration To get the original unction () rom / (), we integrate the unction with respect to /. d c The Basic rules or Integration: - n. d n n. d n c c 4 4 c. d

23 . d 5 5 c 5 5 c. d 0 0. d. d 0 c c d d - - = 8 + c - = d

24 Integration o the eponential unction Y= e () dy / d. e. e d e /. c 5 5e.d /. e. d e 5 c e. d e c Integration o the lograthmatic unction y In dy d / /. d In c 4

25 5 c In d / Integration deinite 5 5 /.. c d a b c a c b c d b a b a c c. d c c

26 ] ( ) = [() + ()] [(0) + (0)] = + = 4 [ ] [ ( ) ( ) ( )] [ ( ) ( ) ( )] ( ) ( ) ( ) ( ) a ( ] = Ln ( + + ) Ln ( ) = Ln 6 Ln ( ) = Ln 6

27 Eercise Find (!) () () (4) (5 ( 7

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