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Differential Calculus: Limits and Derivatives Presentation
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PART 1: LIMITS DIFFERENTIAL CALCULUS DEFINITION Differential calculus is a field in mathematics, which study of how functions change when their inputs change. This rate of change of a function with respect to a chosen input value (the independent variable) is called derivative. LIMIT OF A FUNCTION DEFINITION The LIMIT OF A FUNCTION is the value a function (dependent variable) approaches as the independent variable approaches a value. lim π(π₯) = π(π) π₯→π LIMIT OF A FUNCTION THEOREMS ON LIMITS βLIMIT OF A CONSTANT: lim π = π π₯→π βLIMIT OF A CONSTANT AND A FUNCTION: lim π β π π₯ π₯→π = π β lim [π(π₯)] π₯→π LIMIT OF A FUNCTION THEOREMS ON LIMITS βSUM AND DIFFERENCE OF FUNCTIONS: lim π(π₯) ± π π₯ π₯→π = lim [π(π₯)] ± lim [π(π₯)] π₯→π π₯→π βPRODUCT OF FUNCTIONS: lim π(π₯) β π π₯ π₯→π = lim [π(π₯)] β lim [π(π₯)] π₯→π π₯→π LIMIT OF A FUNCTION THEOREMS ON LIMITS βQUOTIENT OF TWO FUNCTIONS: π π₯ lim π₯→π π π₯ lim π π₯ = π₯→π lim π π₯ π₯→π βLIMIT OF POWER OF A FUNCTION: lim π π₯ π = [lim π π₯ ]π π₯→π π₯→π THEOREMS ON LIMITS INDETERMINATE FORMS If the expression obtained after substituting the limit to a given function does not give enough information to determine the original limit, it is known as an indeterminate form. β 00 β ∞0 β 1∞ 0 β 0 β∞−∞ β0×∞ ∞ β ∞ INDETERMINATE FORMS L’HOSPITALS RULE Generally, the L’Hospitals Rule works only for the limit of a quotient that takes the indeterminate form 0/0 or ∞/∞. π π₯ πΌπ: lim π₯→π π π₯ π π₯ π‘βππ: lim π₯→π π π₯ lim π π₯ 0 ∞ = = ππ lim π π₯ 0 ∞ π₯→π π₯→π π′ π₯ = lim π₯→π π′ π₯ lim π π π₯ π′′ π₯ π₯→π = lim …= π₯→π π′′ π₯ lim ππ π₯ π₯→π LIMITS OF SPECIAL FUNCTIONS L’HOSPITALS RULE sin π₯ β lim =1 π₯→0 π₯ sin π₯ β lim =0 π₯→∞ π₯ β lim cos π₯ = 0 π₯→∞ π₯ β lim tan π₯ = 1 π₯→0 π₯ β 1 − cos π₯ lim =0 π₯→0 π₯ β 1 1+π₯ π₯ β 1 lim 1 + π₯→∞ π₯ π₯ =π lim π₯→0 β π₯ π −1 lim π₯→0 π₯ =π π₯ =1 LIMITS OF SPECIAL FUNCTIONS SPECIAL FUNCTIONS π −1 = ln π β lim π₯→0 π₯ π₯ π − ππ β lim = πππ−1 π₯→0 π₯ − π β lim π₯ = ∞ π₯ β lim = ∞, π ≠ 0 π₯→∞ π π = ∞, π ≠ 0 β lim π₯→0 π₯ π β lim = 0, π ≠ 0 π₯→∞ π₯ π₯ π₯→∞ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 1 Evaluate the limit of the function: π₯2 − 1 lim π₯→1 π₯ − 1 A. 0 B. 1 C. 2 D. ∞ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 2 Evaluate the limit of the function: 3 π₯ − 2π₯ + 5 lim π₯→∞ 2π₯ 3 − 7 A. 0 B. 0.5 C. 2 D. LDE SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 3 Evaluate the limit of the function: sin 3π₯ lim π₯→0 π₯ A. 0 B. 1 C. 3 D. ∞ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 4 Evaluate the limit of the function: lim 2 − π₯ tan ππ₯ 2 π₯→1 A. π 2π B. π π/2 C. π 1/π D. π 2/π SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 5 Evaluate the limit of the function: 4 2 3π₯ − 2π₯ + 7 lim π₯→∞ 5π₯ 3 + π₯ − 3 A. 3/5 B. 0 C. 1 D. ∞ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 6 Evaluate the limit of the function: 100 sin π₯ lim 99 π₯→0 π₯ sin(2π₯) A. 1/2 B. 1/4 C. 1 D. 2 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 7 Evaluate the limit of the function: lim sin(1/π₯) π₯→0 A. 0 B. 1 C. ∞ D. DNE PART 2: DERIVATIVES DERIVATIVES OVERVIEW Given a continuous function y = f(x), the derivative of y or f(x) is defined as the (instantaneous) rate of change of y with respect to the independent variable x, expressed mathematically as: ππ¦ βπ¦ π π₯ + βπ₯ − π π₯ π π₯ = = = lim ππ₯ βπ₯ βπ₯→0 βπ₯ ′ DERIVATIVES OVERVIEW Graphically, the (instantaneous) rate of change of a function is the SLOPE OF THE TANGENT LINE AT A POINT on a graph of that function. MEAN VALUE THEOREM DEFINITION If f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there is a number c in (a,b) such that: π π − π(π) π π = π−π ′ NOTE that the value of ‘c’ must be inside the interval (a,b) ROLLE’S THEOREM DEFINITION If f(x) is continuous over a closed interval [a,b] and differentiable on the open interval (a,b), and if f(a)=f(b)=0, then there is at least one number c in (a,b) such that: π′ π = 0 π π = π(π) NOTE that the value of ‘c’ must be inside the interval (a,b) DERIVATIVES OF FUNCTIONS COMMON DERIVATIVES β ALGEBRAIC FUNCTIONS π 1. π =0 ππ₯ π ππ’ ππ£ 4. π’±π£ = ± ππ₯ ππ₯ ππ₯ π 2. π₯ =1 ππ₯ π ππ£ ππ’ 5. π’π£ = π’ +π£ ππ₯ ππ₯ ππ₯ π π ππ’ π−1 3. π’ = ππ’ ππ₯ ππ₯ ππ’ ππ£ π£ −π’ π π’ 6. = ππ₯ 2 ππ₯ ππ₯ π£ π£ DERIVATIVES OF FUNCTIONS COMMON DERIVATIVES β TRIGONOMETRIC FUNCTIONS π ππ’ 1. sin π’ = cos π’ ππ₯ ππ₯ π ππ’ 2 4. cot π’ = − csc π’ ππ₯ ππ₯ π ππ’ 2. cos π’ = − sin π’ ππ₯ ππ₯ π ππ’ 5. sec π’ = sec π’ tan π’ ππ₯ ππ₯ π ππ’ 2 3. tan π’ = sec π’ ππ₯ ππ₯ π ππ’ 6. csc π’ = − csc π’ cot π’ ππ₯ ππ₯ DERIVATIVES OF FUNCTIONS COMMON DERIVATIVES β INVERSE TRIGONOMETRIC FUNCTIONS π 1 ππ’ −1 1. sin π’ = ππ₯ 1 − π’2 ππ₯ π −1 ππ’ −1 4. cot π’ = ππ₯ 1 + π’2 ππ₯ π −1 ππ’ −1 2. cos π’ = ππ₯ 1 − π’2 ππ₯ π 1 ππ’ −1 5. sec π’ = ππ₯ π’ π’2 − 1 ππ₯ π 1 ππ’ −1 3. tan π’ = ππ₯ 1 + π’2 ππ₯ π −1 ππ’ −1 6. csc π’ = ππ₯ π’ π’2 − 1 ππ₯ DERIVATIVES OF FUNCTIONS COMMON DERIVATIVES β LOGARITHMIC FUNCTIONS π log π π ππ’ 1. log π π’ = ππ₯ π’ ππ₯ π 1 ππ’ 2. ln π’ = ππ₯ π’ ππ₯ ; π ≠ 0, 1 DERIVATIVES OF FUNCTIONS COMMON DERIVATIVES β EXPONENTIAL FUNCTIONS π π’ ππ’ π’ 1. π = π ln π ππ₯ ππ₯ π π’ ππ’ π’ 2. π =π ππ₯ ππ₯ PARTIAL DIFFERENTIATION DEFINITION In evaluating partial derivative of a function with respect to certain independent variable, PERFORM CONVENTIONAL DIFFERENTIATION WITH RESPECT TO THAT VARIABLE and TREAT OTHER VARIABLES CONSTANT. ππ(π₯, π¦, π§) = ππ₯ ππ₯ IMPLICIT DIFFERENTIATION DEFINITION In evaluating derivatives of an implicit function (cannot be expressed as y = f(x)) with respect to the independent variable, PERFORM CONVENTIONAL DIFFERENTIATION WITH RESPECT TO THAT VARIABLE but INCLUDE THE IMPLICIT DERIVATIVE NOTATION. ππ π₯, π¦ ππ¦ ′ = π π₯, π¦ ππ₯ ππ₯ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 8 Find the value of ‘c’ that will satisfy the MVT in the interval (2,5): π₯3 π¦= − 2π₯ 2 + 3π₯ 3 A. 0.268 B. 3.732 C. 0.268, 3.732 D. nothing satisfies SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 9 Find the value of ‘c’ that will satisfy the conditions Rolle’s Theorem in the interval (2,5): π₯3 π¦= − 2π₯ 2 + 3π₯ 3 A. 0 B. 1 C. 3 D. 1 and 3 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 10 Find the derivative of the function: π₯+1 π¦= π₯+2 2 A. π₯+2 2 π₯ B. π₯+2 2 1 C. 2 π₯ +4π₯+4 D. none of the above SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 11 Find the derivative of the function: π¦= A. B. −6π₯ 2−3π₯ 2 −3π₯ 2−3π₯ 2 2 − 3π₯ 2 C. D. 6π₯ 2−3π₯ 2 3π₯ 2−3π₯ 2 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 12 Find the derivative of the function: π¦ = cos(2π₯ − 3) A. −sin(2π₯ − 3) B. sin(2π₯ − 3) C. −2sin(2π₯ − 3) D. 2sin(2π₯ − 3) SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 13 Find the derivative of the function: π¦ = arctan(3π₯) 1 A. 2 6π₯ +1 1 C. 2 9π₯ +1 3 B. 2 6π₯ +1 3 D. 2 9π₯ +1 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 14 Find the derivative of the function: π₯ π¦ = arcsinh 2 A. B. 1 π₯ 2 +2 1 π₯ 2 +4 C. D. 2 π₯ 2 +2 2 π₯ 2 +4 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 15 Find dy/dx of the function at 7, 3 π₯ 2 + 4π¦ 2 − 10π₯ − 16π¦ + 5 = 0 A. –0.5 B. 0.5 C. 2 D. none of the above SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 16 Find the second derivative of the y with respect to w: π¦ = (3π€ 2 − 4)(3π€ 2 + 4) A. 108π€ B. 108π€ 2 C. 54π€ D. 54π€ 2 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 17 Find ππ₯ of the function: π π₯, π¦ = 2π₯ 2 π¦ + π₯π¦ 2 A. 2π₯ 2 + 2π₯π¦ 2 B. 4π₯π¦ + π¦ C. 4π₯π¦ + 2π₯π¦ D. none of the above SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 18 Find ππ₯π¦ of the function: π π₯, π¦, π§ = π₯ 2 cos(π§π¦ 3 ) + π§ sinh(2π₯π¦) A. −6π₯π§π¦ 2 sin π§π¦ 3 + 2π§ 2π₯π¦ sinh 2π₯π¦ + cosh 2π₯π¦ 2 3 B. 6π₯π§π¦ sin π§π¦ − 2π§ 2π₯π¦ sinh 2π₯π¦ + cosh 2π₯π¦ C. −6π₯π§π¦ 2 sin π§π¦ 3 + 2π§ 2π₯π¦ sinh 2π₯π¦ − cosh 2π₯π¦ D. 6π₯π§π¦ 2 sin π§π¦ 3 − 2π§ 2π₯π¦ sinh 2π₯π¦ − cosh 2π₯π¦ PART 3: DERIVATIVE APPLICATIONS SLOPE OF A CURVE DEFINITION The slope of the curve at any point can be calculated by obtaining the FIRST DERIVATIVE of the curve which is also the slope of the line TANGENT at any point. π¦2 − π¦1 βπ¦ ππ¦ π= = = = π¦′ π₯2 − π₯1 βπ₯ ππ₯ SLOPE OF A CURVE NORMAL LINE SLOPE The slope of the line normal to a curve at any point is the negative reciprocal of the slope of the line tangent to the curve. 1 ′ π¦ππππππ = πππππππ = − ′ π¦ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 19 Find the slope of the curve at (1,2): π₯ 2 + 4π₯π¦ + π¦ 2 + 3π₯ − 5π¦ − 6 = 0 A. 10/3 B. -13/3 C. 13/3 D. none of the above SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 20 Find the equation of the line normal to the curve at (1, –2): π¦ 2 = 5π₯ − 1 A. 4π₯ − 5π¦ − 14 = 0 B. 5π₯ − 4π¦ + 8 = 0 C. 4π₯ + 5π¦ − 14 = 0 D. 5π₯ + 4π¦ + 8 = 0 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 21 Find the angle between the curve π¦ = π₯ 2 and the line π¦ = π₯ + 12 at the point of their intersection, (4, 16). A. 32o B. 35o C. 38o D. 42o RADIUS OF CURVATURE FORMULA The radius of curvature of a curve at any point can be computed as: π= 1+ 3 π¦′ 2 2 π¦′′ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 22 Find the approximate radius of curvature of the function at (1, –1): 2 π¦ = π₯ − 3π₯ + 1 A. 2 B. 3 C. 2 2 D. 2 5 CRITICAL POINTS OF A CURVE CRITICAL POINTS A curve has 3 critical points: β MINIMUM POINT – the point where the FIRST DERIVATIVE of the function is ZERO and SECOND DERIVATIVE is POSITIVE. β MAXIMUM POINT – the point where the FIRST DERIVATIVE of the function is ZERO and SECOND DERIVATIVE is NEGATIVE. CRITICAL POINTS OF A CURVE CRITICAL POINTS A curve has 3 critical points: β POINT OF INFLECTION – the point where the SECOND DERIVATIVE of the function is ZERO. MAXIMA AND MINIMA SOLVING MAXIMA / MINIMA PROBLEMS Generally, maxima and minima solves for the value of a certain parameters (independent) that will produce the maximum or minimum value of a dependent parameters such as length, area, volume, time, etc. MAXIMA AND MINIMA SOLVING MAXIMA / MINIMA PROBLEMS In solving for Maximum/Minima problems, remember these steps: i. Determine the variable to be maximize/minimize. ii. Express the variable to be maximize/minimize in terms of one independent variable using relations. iii. Find the derivative of the variable to be maximize/minimize with respect to the independent variable. MAXIMA AND MINIMA SOLVING MAXIMA / MINIMA PROBLEMS In solving for Maximum/Minima problems, remember these steps: iv. Equate the derivative to zero and solve for the values of the independent variable and unknown. v. Check the values if it will produce the maximum or minimum using 2nd derivative. y’’ = + (minimum) y‘’ = – (maximum) SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 23 Determine the minimum, maximum and inflection of the function respectively. π¦ = π₯ 3 − 9π₯ 2 + 15π₯ − 3 A. 5, −28 , 1, 4 , 3, −12 B. 5, −28 , 3, −12 , 1, 4 C. 1, 4 , 5, −28 , 3, −12 D. 1, 4 , 3, −12 , 5, −28 SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 24 What is the area of the largest rectangle that can be inscribed in a semi circle of radius 10 cm? SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 25 The sum of two numbers is 21. The product of one of the numbers by the square of the other is to maximum, what are the numbers? SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 26 A square piece of tin of side 24 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the size of the square to be cut off so that the volume of the box is maximum? SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS SOLUTION NO. 26 TIME RATES SOLVING TIME RATES In solving for time rates problems, remember these steps: i. Find the equation in terms of the given and unknown variables. ii. Take the derivative of the function with respect to TIME. iii. Substitute the given constants and rates and solve for the unknown rates. SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 27 A kite, at a height of 60 ft is moving horizontally at a rate of 5 ft/s away from the boy who flies it. How fast is the cord being released when there are 100 ft out? SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS SOLUTION NO. 27 π¦ = 60 ππ‘ ππ₯ = 5 ππ‘/π ππ‘ SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS QUESTION NO. 28 Water is flowing into a conical reservoir 20 ft deep and 10 ft across the top, at a rate 15 cu. ft. per minute. Find how fast is the water rising when it is 8 ft deep? SAMPLE PROBLEMS IN DIFFERENTIAL CALCULUS SOLUTION NO. 28 THANK YOU!
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