Calculus Differentiation Norhafizah Md Sarif Faculty of Industrial Science & Technology

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1 Calculus Differentiation By Norhafizah Md Sarif Faculty of Industrial Science & Technology

2 Description Aims This chapter is aimed to : 1. introduce the concept of integration. evaluate the definite and indefinite integral 3. eplain the basic properties of integral 4. compute the integral using different techniques of integration Epected Outcomes 1. Students should be able to describe the concept of antiderivatives. Students should be able to eplain about indefinite integral and definite integral 3. Students should be able to know the basic properties of definite integrals 4. Student should be able to determine the appropriate techniques to solve difficult integral. References 1. Abdul Wahid Md Raji, Hamisan Rahmat, Ismail Kamis, Mohd Nor Mohamad, Ong Chee Tiong. The First Course of Calculus for Science & Engineering Students, Second Edition, UTM 016.

3 Content 1 3 Introduction to Differentiation Derivative of a Function Techniques of Differentiation

4 Applications Do you know that we can use differentiation to find the highest point and the lowest point of the roller coaster track? highest point lowest point Classes/CalcI/Tangents_Rates.as p

5 We use the derivative to determine the maimum and minimum values of particular functions - cost, strength, - amount of material used in a building - profit & loss

6 Introduction to Differentiation Differentiation is all about finding rates of change of one quantity compared to another. The process of finding a derivative is called differentiation The most common notations for the derivative are y ' dy d f '( )

7 Derivative of a Function Definition First Principle: The derivative of the function f( ) with respect at a point ( ) if the function where f( ) lim h0 f ( h) f ( ) h The process of finding derivative using the above definition is called the first principle.

8 Differentiation using first principle 1 Write an epression for Write an epression for f( ) f ( h) 3 4 Substitute f( ) and f ( h) into the formula Simplify the epression 5 Evaluate limit lim h0 f ( h) f ( ) h

9 Eample By using differentiation from the first principle, find the derivatives of the following functions 1 (a) 4 (b) 5 (c) (d) (a) Given f( ) 4, then f ( h) 4 Substitute f( ) and f ( h) into first principle formula, f So, 4 4 '( ) lim h0 h f '( ) 0

10 Eample (b) f ( ) 5 f ( ) 5 1. f ( h) h 5 f ( h) 5 5 '( ) lim h0 h h f '( ) lim lim h0 h h0

11 Eample (c) f( ) f( ) f ( h) h 1. f f '( ) 1 1 lim h h 0 h ( h) ( h) h '( ) lim lim h0 h h0 h ( h ) 1 1 f '( ) lim h0 ( ) h

12 (d) f ( ) f ( h) h f '( ) lim h0 lim h0 h h h h h h ( h ) ( ) lim h0 h ( h ) 1 1

13 Common Functions Functions Derivatives dy Constant y c 0 d Power Eponential Logarithms Trigonometry ( is in rad) Hyperbolic Inverse Trigonometry n dy n1 y n d dy y e e d dy y a a (ln a) d y ln dy 1 d y log a dy 1 d ln a y sin dy cos d y cos dy sin d y tan dy sec d y sec dy sec tan d y sinh dy cosh d y cosh dy sinh d y tanh y dy sech d dy 1 d 1 dy 1 d 1 dy 1 d 1 1 sin y cos y 1 1 tan

14 Techniques of Differentiation Definition Chain Rule: If y f ( u) and u g( ), then dy d dy du du d The process of finding derivative using the above definition is called the chain rule. t Basic function such as sin t, e, ln t can be solve using table of 3t differentiation. However, for case sin t, e, ln5t we need to use chain rule to obtain the solution.

15 Eample Differentiate the following functions 4 3 (a) y (3 5), (b) y 1 t, (c) y 5sin( ), (a) and 3 5, hence y u y u dy 4 du Then, 5 u, and 3 du d dy dy du (5 u )(3) 15u 15(3 5) d du d (b) y 1 t and u 1 t, hence y u 1/ 1 dy 1 du Then, u, and t du d 1 1 dy dy du 1 t ( u )( t) tu d du d 1 t

16 How about these eamples? (a) y (4 1) (3 ) (b) y sin (c) y t t e t (d) y e cos(3 t) (e) y 3 sin (f) y (g) y 4e 1 15

17 Techniques of Differentiation Product rule is a formula used to find the derivatives of product of two or more functions Definition Product Rule: If u ( ) and v ( ) are differentiable function and y u( ) v( ), then dy dy dv v u d du d 16

18 Eample Differentiate the following functions t (a) (4 1)(3 ), (b) cos(3 ), (c) sin y y e t y (a) dy d d (4 1) (3 ) (4 1) (3 ) d d d (41) (3) (4) (3 ) 4 3 (b) dy t d d t e (cos3 t) ( e ) cos(3 t) dt d d t t e ( 3sin 3 t) ( e ) cos(3 t) t t 3 sin 3 cos3 e t e t

19 Techniques of Differentiation The quotient rule is a formal rule for differentiating problems where one function is divided by another Definition Quotient Rule: If u ( ) and v ( ) are differentiable u ( ) function and y where v ( ) 0, then v ( ) du dv v u dy d d d v 16

20 Eample Differentiate the following functions 4 1 sin 4e (a) y, (b) y, (c) y 3 1 (a) d d (3 ) (4 1) dy d d d 3 3 (4) (b) d d sin ( ) (sin ) dy d d d (cos ) 1sin cos sin

21 Conclusion #1 Any indefinite integral will have +c at the end of the solution. There are two approach in getting the solution of definite integral: by changing the limit of into u, or by changing function u into and use the original limit. Product or quotient function cannot be integrate directly. Appropriate techniques should be used to solve this type of integral In integration by substitution, making appropriate choices for u will come with eperience. Selecting u for by part techniques should follow the LATE guideline. If the power of denominator is less than the power of numerator, then the fraction is called proper fraction

22 Author Information Norhafizah Binti Md Sarif Lecturer Faculty of Industrial Sciences & Technology (website) Universiti Malaysia Pahang Google Scholar: : Norhafizah Md Sarif Scopus ID : UmpIR ID: 3479