diff cal

MATH 401 – DIFFERENTIAL CALCULUS 8. A photographer is taking a picture of a painting hung in an art gallery. The height of a painting is 4 ft. The camera lens is 1 ft below the lower edge of the painting. How far should the camera be from the painting to maximize the angle subtended by the lens camera? Ans. 2.24 ft Find two positive numbers that satisfy the given requirements. 9. The second number is the reciprocal of the first number and the sum is minimum. Ans. 1 10.The sum of the first number squared and the second number is 54 and the product is a maximum. Ans. 3 and 36 198

MATH 401 – DIFFERENTIAL CALCULUS 4.5 RELATED RATES A problem in related rates is one involving rates of change of related variables. In real-world applications involving relates rates, the variables have a specific relationship for values of time t. This relationship usually expressed in the form of an equation which represents a mathematical model of the situation. The table below lists examples of mathematical models involving rates of change. Verbal Statement Mathematical Model The velocity of a car after traveling for when 1 hour is 120 kilometers per hour. Water is being pumped into a swimming pool at a rate of 10, 000 L per hour A wheel is revolving at a rate of 50 revolutions per minute. An electric charge is flowing in a wire at a rate of 2 Coulombs per second. Here are the guidelines for solving Related-Rate Problems GUIDELINES FOR SOLVING RELATED-RATE PROBLEMS 1. GIVEN. Identify all given quantities and quantities to be determined. Make a sketch and label the quantities. 2. MODEL. Write an equation involving the variables whose rates of change either are given or are to be determined. This is a mathematical model of the situation. 3. CALCULUS. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 4. SIMPLIFY. Substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change. 5. CONCLUSION. Write a conclusion, consisting of one or more complete sentences, that answers the questions of the problem. Be sure your conclusion contains the correct units of measurement. Note: When using these guidelines, be sure to perform Step 3 before Step 4. Substituting the known values of the variables before differentiating will produce an inappropriate derivative. 199

MATH 401 – DIFFERENTIAL CALCULUS 4.5.1. Related-Rate problems involving algebraic functions Example 4.5.1.1 Ladder Problem A 20-ft ladder leaning against a vertical wall starts to slide. The bottom of the ladder is pulled horizontally away from the wall at 4 ft/s. Determine how fast the top of the ladder is sliding down the wall when the bottom is 12 ft from the wall. Solution: Step 1: GIVEN Draw the figure and label all quantities 20 ft y x Figure 1 . Ladder leaning against a wall t = time that has elapsed since the ladder started to slide down the wall x = distance from the bottom of the ladder to the wall at time t y = distance from the ground to the top of the ladder at time t rate at which the bottom of the ladder moves horizontally from the wall rate at which the top of the ladder slides down the wall Step 2: MODEL Because the bottom of the ladder is pulled horizontally away from the wall at 4 ft/s, . In order to determine when 12, write an equation to relate x and y. From the Pythagorean Theorem, . Solving for y in terms of x yields Step 3: CALCULUS Using the Chain Rule, implicitly differentiate both sides with respect to t, 200

MATH 401 – DIFFERENTIAL CALCULUS Step 4: SIMPLIFY Substitute the known values of x, y, and in the above equation and solve for . When 12, Because , The minus sign indicates that y decreases as t increases. Step5: CONCLUSION The top of the ladder is sliding down the wall at the rate of 3 ft/s when the bottom of the ladder is 12 ft from the wall Example 4.5.1.2 An Inflating Balloon A spherical balloon is being inflated at a rate of 10 ft3/min. How fast the radius of the balloon is expanding when the radius is 1 foot. Solution: Step 1: GIVEN Draw the figure and label all quantities Figure 2. Spherical Balloon Step 2: MODEL We need an equation which relates V and r for a sphere; that is the volume of the sphere 201

MATH 401 – DIFFERENTIAL CALCULUS Step 3: CALCULUS Using the Chain Rule, implicitly differentiate both sides with respect to t, Step 4: SIMPLIFY in the above equation, we have . Substitute the known values of r and The positive sign indicates that r increases as t increases. Step5: CONCLUSION When the radius of the balloon is 1 ft, it is expanding at a rate of approximately 0.8 ft3/min. Example 4.5.1.3 Flow Rate in a Cylindrical Tank Water is flowing into a 4-ft radius vertical cylindrical tank at the rate of 24ft3/min. How fast is the surface of water rising? Solution: Step 1: GIVEN Draw the figure and label all quantities h = level of water in the tank r = radius of tank = 4 Figure 3. Cylindrical Tank 202

MATH 401 – DIFFERENTIAL CALCULUS Step 2: MODEL We need an equation which relates V and h ; that is the volume of a cylinder Step 3: CALCULUS Using the Chain Rule, implicitly differentiate both sides with respect to t, Step 4: SIMPLIFY in the above equation, we have . Substitute the known value of The positive sign indicates that h increases as t increases. Step5: CONCLUSION When the radius of the tank is 4 ft, the water surface is rising at a rate of approximately 0.48 ft/min. Example 4.5.1.4 Flow Rate in a Conical Tank A water tank in the shape of an inverted cone is leaking water at a rate of 4 ft3/hour. The base radius of the tank is 5 ft and the height of the tank is 15 ft. When the depth of water is 6 ft, (a) at what rate is the depth of water in the tank changing, (b) at what rate is the radius of the top of water in the tank changing? Solution: h = depth of water in the Step 1: GIVEN tank Draw the figure and label all quantities r = radius of top of water in the tank 203

MATH 401 – DIFFERENTIAL CALCULUS Figure 4. Conical Tank Step 2: MODEL We need an equation which relates V , r and h ; that is the volume of water in the tank at any time To answer (a) we need to express V in terms of h only. To do this, we need a secondary equation that will relate r and h. Consider two similar triangles, the ratios of any two sides are equal. From the figure, we have Substitute this into the volume formula gives Step 3: CALCULUS Using Chain Rule, implicitly differentiate both sides with respect to t, Step 4: SIMPLIFY Substitute the known value of (negative because water is leaking out), and the depth h = 6 in the above equation, we have The negative sign indicates that h decreases as t increases. To answer part (b), use the equation of r in terms of h, . Differentiate both sides with respect to time gives . Substituting the value of , we have The negative sign indicates that r decreases as t increases. 204

MATH 401 – DIFFERENTIAL CALCULUS Step5: CONCLUSION When the depth of water in the tank is 6 ft, (a) the depth of water in the tank decreases at a rate approximately 0.28 ft/hour ; and (b) the radius of the top of water in the tank decreases at a rate of approximately 0.10 ft/hour. Example 4.5.1.5 Flow Rate in a Trapezoidal Trough A trapezoidal trough is 10 ft long, 4 ft wide at the top, 2 ft wide at the bottom and 2 ft deep. If water flows in at a rate of 10 ft3/min, find how fast the water surface is rising, when the water is 1 ft deep. Solution: Step 1: GIVEN Draw the figure and label all quantities y = depth of water in the trough For a trapezoid: b1 = 2 b2 = x +2+x = 2x +2 h =y Figure 5. Water flowing in a trapezoidal trough Step 2: MODEL We need an equation which relates V and y ; that is the volume of water in the trough at any time. Recall volume of a container is , with the area of a trapezoid . The volume of water in the trapezoidal trough at any time is . We need to express this volume V in terms of y only. To do this, 205

MATH 401 – DIFFERENTIAL CALCULUS express x in terms of y. Consider two similar triangles, the ratios of any two sides are equal. From the figure, we have Substitute this into the volume formula gives Step 3: CALCULUS Using Chain Rule, implicitly differentiate both sides with respect to t, Step 4: SIMPLIFY Substitute the known value of (positive because water flows in), and the depth y = 1 in the above equation, we have The positive sign indicates that y increases as t increases. Step5: CONCLUSION When the depth of water in the tank is 1 ft, the water surface is rising at a rate of 1/3 ft/min. 4.5.2 Related Rates Problems involving Transcendental Functions Example 4.5.2.1 Rate of Separation Two people are 50 ft apart. One of them starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min. At what rate is distance between the two people changing when ? 206

MATH 401 – DIFFERENTIAL CALCULUS Moving Person x θ Stationary Person 50ft Figure 6 Solution: Step 1: GIVEN Let x be the distance between the two person at any given time. Step 2: MODEL We need an equation which relates x and θ. From the figure, consider a right triangle, apply SOHCAHTOA, we have Step 3: CALCULUS Using Chain Rule, implicitly differentiate both sides with respect to t, Step 4: SIMPLIFY and in the above equation, Substitute the known value of we have The positive sign indicates that x increases as t increases. Step5: CONCLUSION When , the rate at which the distance between two people is changing is approximately 0.71 ft/min. 207

MATH 401 – DIFFERENTIAL CALCULUS Example 4.5.2.2 Changing Angle of Elevation A camera is on the ground, filming a rocket launch. The rocket is rising according to the position function , where y is measured in feet and t in seconds. Find the rate of change of the camera’s angle at 1 seconds after the rocket initially launches. The camera is 1,000 feet away from the rocket. Solution: Step 1: GIVEN Draw the figure and label all the quantities: Figure 7. Rocket Launching Step 2: MODEL We need an equation which relates s and θ. From the figure, consider a right triangle, apply SOHCAHTOA, we have Step 3: CALCULUS Using Chain Rule, implicitly differentiate both sides with respect to t, Differentiating the function with respect to t gives . To solve for , consider the figure, and apply Pythagorean theorem and SOHCAHTOA, we have 208

MATH 401 – DIFFERENTIAL CALCULUS Step 4: SIMPLIFY , we have Substitute the known values of and Then The positive sign indicates that θ increases as t increases. Step5: CONCLUSION The rate of change of the camera’s angle at 1 seconds after the rocket initially launches is approximately 0.02 rad/s. Example 4.5.2.3 Ladder Problem A 20-ft ladder leaning against a vertical wall starts to slide. If the top of the ladder is slides down at 4 ft/s. How fast is the angle of elevation of the ladder changing when the lower end of the ladder is 12 ft from the wall. Solution: Step 1: GIVEN Draw the figure and label all quantities 20 ft y θ x Figure 8 . Ladder leaning against a wall t = time that has elapsed since the ladder started to slide down the wall x = distance from the bottom of the ladder to the wall at time t y = distance from the ground to the top of the ladder at time t angle of elevation rate at which the top of the ladder slides down the wall 209

MATH 401 – DIFFERENTIAL CALCULUS Step 2: MODEL In order to determine when 12, write an equation to relate y and θ. From the figure, consider the right triangle and used SOHCAHTOA Step 3: CALCULUS Differentiate both sides with respect to t, Step 4: SIMPLIFY Substitute the known values of x, y, and in the above equation and solve for . When 12, Because , The minus sign indicates that θ decreases as t increases. Step5: CONCLUSION when the lower The angle of elevation of the ladder decreases by end of the ladder is 12 ft from the wall. 210

MATH 401 – DIFFERENTIAL CALCULUS 4.5.3. Motion Problems In this section, we will consider the application of the derivative of a function in analyzing the motion of a particle on a line. Such motion is called rectilinear motion. Let s be the directed distance(displacement) of the particle from the origin O at time t. Then f is the function defined by the equation The instantaneous velocity of the particle at t units of time is given as The instantaneous rate of change of velocity is called the instantaneous acceleration given as or Example 4.5.3.1 A particle is moving along a horizontal line according to the equation Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine the instant when the particle is at rest. What is the acceleration of the particle at t = 1 s? Solution: To determine the instantaneous velocity, differentiate s with respect to t, 211

MATH 401 – DIFFERENTIAL CALCULUS Determine the sign of various v for various intervals of t, and the results are given below: t-2 t–6 V Conclusion - - + particle is moving to the right 0 - 0 particle is at rest + - - particle is moving to the left + 0 0 particle is at rest + + + particle is moving to the right For the instantaneous acceleration, differentiate the velocity v with respect to time t, At t = 1 s, Since at t = 1 s, the particle is moving to the right, the negative acceleration indicates that its velocity is decreasing. Example 4.5.3.2 A ball is thrown vertically upward from the ground with an initial velocity of 64 ft/s. The equation of motion is given by (a) How high the ball go and how many seconds it takes for the ball to reach its highest point? (b)Find the instantaneous velocity of the ball at 1 s and 3 s. (c) Find the instantaneous velocity of the ball when it reaches the ground. (d)What is the acceleration of the ball at any time t? Solution: (a)At the highest point, the instantaneous velocity is zero, that is Differentiate s with respect to t, then set 212

MATH 401 – DIFFERENTIAL CALCULUS Substitute this t into the equation of motion s, we have Therefore, the ball reaches a highest point of 64 ft above the starting point at 2 s. (b) The instantaneous velocity of the ball: At the end of 1 s, the ball is rising with an instantaneous velocity of 32 ft/s. At the end of 3 s, the ball is falling with an instantaneous velocity of -32 ft/s. (c) Since the ball reaches the highest point at t = 2 s, then it takes a total of 4 s to reach the ground. Substituting it to the instantaneous velocity, we have At the end of 4 s, the ball will reach the ground with an instantaneous velocity of -64 ft/s. (e) The instantaneous acceleration of the ball is It means that the particle is constantly accelerating at 32 ft/s2 downward. That -32 ft/s2 is also the acceleration due to gravity. Example 4.5.3.3 A particle is moving along a line according to the equation of motion , with . Find the value of t for which the measure velocity is (a) 0, (b) 1; and (c) 2. Solution: To determine the instantaneous velocity, differentiate s with respect to t, 213

MATH 401 – DIFFERENTIAL CALCULUS (a) When v = 0 (b)When v = 1 Since there is no negative time t, we only consider . (c) When v = 2 There is no time t possible when v = 2. Example 4.5.3.3 A particle is moving on a line according to the equation of motion where s centimeters is the directed distance of the particle from the origin at t seconds. Express the velocity and acceleration as a function of time t. Solution: To determine the velocity of a particle as a function of time t, differentiate s with respect to t. To determine the acceleration of a particle as a function of time t, differentiate v with respect to t. 214

MATH 401 – DIFFERENTIAL CALCULUS Comparing the acceleration and the directed distance s, we have Since the acceleration and the position s are oppositely directed, then the motion is called simple harmonic. Example 4.5.3.4 A C800 airplane take off from an airport at sea level and its altitude (in feet) and time t (in minutes) is given by What is the rate of climb at t = 3min? Solution: To determine the rate of climb (vertical velocity) of the airplane, differentiate h with respect to t, we have At t = 3 min, 215

MATH 401 – DIFFERENTIAL CALCULUS Example 4.5.3.5 A particle is moving on a line according to the equation of motion (3 sinh t + 4 cosh t) where s centimeters is the directed distance of the particle from the origin at t seconds. Express the velocity and acceleration as a function of time t. Solution: To determine the velocity of a particle as a function of time t, differentiate s with respect to t. To determine the acceleration of a particle as a function of time t, differentiate v with respect to t. 216

MATH 401 – DIFFERENTIAL CALCULUS Exercises 4.5 Solve the following related-rate and motion problems: 1. Water is flowing at the rate of 2 m3/min into a tank in the form of an inverted cone having an altitude of 16 m and a radius of 4 m. How fast is the water level rising when the water is 5 m deep? Ans. 40.74 cm/min 2. Two cars, one is going due East at the rate of 90km/hr and the other is going due South at the rate of 60 km/hr, are traveling toward the intersection of two roads. At what rate are the cars approaching each other at the instant when the first car is 0.2 km and the second car is 0.15 km from the intersection? Ans. 108 km/hr 3. If a body of weight W pounds is dragged along a horizontal floor at constant velocity by means of a force of magnitude F pounds and directed at an angle of radians with the plane of the floor, then F is given by the equation where is a constant called the coefficient of friction. If , find the Ans. 2W instantaneous change of F with respect to θ when . 4. Boyle’s Law for the expansion of gas is , where P is the number of pounds per square unit of pressure, V is the number of cubic units of volume of a gas, and C is a constant. At a certain instant the pressure is 3000 lb/ft2, and the volume is 5 ft3, and the volume is increasing at the rate of 3 ft3/min. Find the rate of change of pressure at this instant. Ans. 1,800lb/ft2 per min 5. If P pounds per square foot is the atmospheric pressure at a height of h feet above sea level, then Find the time rate of change of the atmospheric pressure outside an airplane that is 5,000 ft high and rising at the rate of 160 ft/s? Ans. -9.17lb/ft2 per second 6. A picture 40 cm high is placed on a wall with its based 30 cm above the level of the eye of an observer. If the observer is approaching the wall at the rate of 40 cm/s, how fast is the measure of the angle subtended at the 217

MATH 401 – DIFFERENTIAL CALCULUS observer’s eye by the picture changing when the observer is 1 m from the wall? Ans. 0.08 rad/s 7. Suppose we have two resistors with resistances and measured in ohms(Ω) connected in parallel. The equivalent resistance, R, iis then given by Supposed is increasing at a rate of 0.4Ω/min and is decreasing at a rate of 0.7Ω/min . At what rate is R changing when and ? Ans. -0.00 Ω/min 8. A man 6 feet tall walks at a rate of 5 feet per second away from the light that is 15 feet above the ground. When he is 10 feet from the base of the light, (a) At what rate is the tip of his shadow moving? Ans. (b)At what rate is the length of his shadow moving? Ans. 9. A particle is moving along a line according to the equation of motion Where s meters is the directed distance of the particle from the origin at t seconds. Find the t, s, and velocity v when the acceleration a = 0. Ans. 10. A stone is dropped from a building 256 ft high. How long it takes the stone to reach the ground? What is the velocity of the stone when it reaches the ground? Ans. Problem Set No 4. Solve the following problems 1. Given y 1 , find df. x 1 2. Given that f x  1 x  2x2 compute df and f when x  1 and x  0.02 . 3. Find the actual and approximate volume of a spherical shell whose inner radius is 3cm and whose thickness is 1/8cm. 218

MATH 401 – DIFFERENTIAL CALCULUS 4. How much variation dr in the radius of a coin can be tolerated if the volume of the coin is to be within 11000 of their ideal volume? 5. Given f x  1 x4  2 x3  2x2 : (a) find the relative extrema of the 23 function using the second-derivative test; (b) determine the values of x at which the relative extrema occur; (c) determine the intervals on which the function is increasing and decreasing; (d) concave upward or concave downward; (e) Locate inflection points if any and (f) Sketch the graph of the function from your answers. 6. A piece of wire 10 ft long is cut into two piece. One piece is bent into the shape of an equilateral triangle and the other piece is bent into a shape of a square. How should a wire be cut so that (a) the combined area of the two figures is as small as possible; (b) the combined area of the two figures is as large as possible? 7. A water tank in the form of an inverted cone is being emptied at the rate of 6m3/min. The altitude of the cone is 24 m, and the radius is 12 m. Find how fast the water level is lowering when the water is 10 m deep? 8. A right-circular cone is to be inscribed in a sphere of given radius. Find the ratio of the altitude to the base radius of the cone of largest possible volume. 9. Find two positive numbers such that their sum is 12 and the sum of their squares is a minimum. 10.A ball is thrown vertically upward from the ground with an initial velocity of 50 m/s. The equation of motion is given by a. How high the ball go and how many seconds it takes for the ball to reach its highest point? b. Find the instantaneous velocity of the ball at 1 s and 3 s. c. Find the instantaneous velocity of the ball when it reaches the ground. d. What is the acceleration of the ball at any time t? 219

MATH 401 – DIFFERENTIAL CALCULUS CHAPTER V PARTIAL DIFFERENTIATION The concept of a function of a single variable is extended to a multi – valued functions. In this chapter, we shall generalize the notion of function of a function expressed in two or more independent variable. For instance, the function f defined by f x, y, z  x2  y2  z 2  25 is a multi – valued function with x, y and z as independent variables. To facilitate learning on multi – valued functions, this chapter presents the definition of derivatives of multi – valued function also known as partial differentiation. Also, chain rule of partial differentiation will be discussed on this chapter. Definitions, theorems an some examples and illustrations presented in this chapter are taken from [1] Larson, R. (2018), [2] Leithold, L. (2002) and [3] Stewart, J. (2016) At the end of this chapter, the learners might be able to: 1. Determine the partial derivatives of functions of two or more variables; and 2. Evaluate the higher-order partial derivatives of functions of two or more variables. 5.1. DEFINITION OF PARTIAL DERIVATIVES OF A FUNCTION The process of finding the partial derivatives of a function is called partial differentiation. Let f be a function in terms of two variables say x and y written as f x, y, then the partial derivatives of f are denoted by the following: Notations Meaning D1 f read as “D sub 1 of f ” and f1 read as “ f sub 1” D1 f , f1 , fx , f denote the partial derivative of f with respect to the x first variable. In the case of f x, y, that is with respect D2 f , f2, fy , f y to x f x read as “ f sub x ” denotes the partial derivative of f with respect to x . f is read as “partial derivative of f with respect to x ” x D2 f read as “D sub 2 of f ” and f2 read as “ f sub ” denote the partial derivative of f with respect to the second variable. In the case of f x, y, that is with respect to y f y read as “ f sub y ” denotes the partial derivative of f 220

MATH 401 – DIFFERENTIAL CALCULUS with respect to y . f is read as “partial derivative of f with respect to y ” y These notations can be used interchangeably to denote partial derivatives of a function. Definition 5.1 gives the formal definition of the partial derivative of a multi– valued function. Definition 5.1. [2] Let f be a function of two variables, x and y . The partial derivative of f with respect to x is that function, denoted by D1 f , such that its function value at any point x, y in the domain of f is given by D1 f x, y  lim f x  x, y  f x, y x0 x if this limit exists. Similarly, the partial derivative of f with respect to y is that function, denoted by D2 f , such that its function value at any point x, y in the domain of f is given by D2 f x, y  lim f x, y  y  f x, y y0 y if this limit exists. Example 5.1. Apply the definition of partial differentiation to find D1 f and D2 f if f x, y  2x2  3xy  y2 . Solution:    D1 f x, y  fx  lim 2x  x2  3x  xy  y2  2x 2  3xy  y 2 . Simplifying, x0 x       lim 2 x2  2xx  x2  3xy  3xy  y2  2x2  3xy  y2 x0 x  lim 2x2  4xx  2x2  3xy  3xy  y 2  2x2  3xy  y 2 x0 x  lim 4xx  2x2  3xy x0 x 221

MATH 401 – DIFFERENTIAL CALCULUS  lim x4x  2x  3y x0 x  lim 4x  2x  3y Applying the theorem on limits x0  4x  20  3y D1 f x, y  f x  4x  3y  D2 f x, y   2x2  3xy  y 2 . Simplifying, fy  lim 2x2  3xy  y  y  y2 y0 y     lim 2x2  3xy  3xy  y 2  2yy  y2  2x2  3xy  y 2 y0 y  lim 2x2  3xy  3xy  y 2  2 yy  y2  2x2  3xy  y 2 y0 y  lim  3xy  2yy  y2 y0 y  lim y 3x  2y  y y0 y  lim  3x  2y  y Applying the theorem on limits y0  3x  2y  0  D2 f x, y  f y  3x  2y Example 5.2. Let f be a function defined by f x, y  x  2y , find f and f . x y Solution: By definition 5.1. we have f  lim x  x  2y  x  2y x x0 x x  x  2 y  x  2 y x  x  2 y  x  2 y  lim  x0 x x  x  2 y  x  2 y   lim x  x  2y  x  2y x0 x x  x  2 y  x  2 y 222

MATH 401 – DIFFERENTIAL CALCULUS x   lim x0 x x  x  2 y  x  2 y  lim 1 x0 x  x  2 y  x  2 y  1 1 x  0  2y  x  2y 2 x  2y  f  1 x 2 x  2 y f  lim x  2y  y  x  2y y y0 y x  2y  y  x  2y x  2y  y  x  2y  lim  x  2y  y  x  2y y0 y x  2 y  2y  x  2y x  2y  y  y x2  lim y y0 2y x  2y  y    lim y0 y x  2y  lim 2 x  2y y0 x  2y  y   2 2 x  2y 2 x  2y x  2y  0   f  1 y x  2 y We shall now extend the concept of Definition 1.1 into a function of n  variables. 223

MATH 401 – DIFFERENTIAL CALCULUS Definition 5.2. [2] Let Px1, x2 ,...xn1, xn  be a point in Rn , and let f be a function of n  variables, x1, x2 ,...xn1, xn . The partial derivative of f with respect to xk is that function, denoted by Dk f , such that its function value at any point P in the domain of f is given by Dk f x1, x2 ,...xn1, xn   lim f x1, x2 ,...xn1, xn  xk ,...xn   f x1, x2 ,...xn1, xn  x0 xk if this limit exists. For instance, if f is a function in three variables x, y and z , then the partial derivatives of f are given by D1 f x, y, z  f  lim f x  x, y, z  f x, y, z x x0 x D2 f x, y, z  f  lim f x, y  y, z  f x, y, z y y0 y D3 f x, y, z  f  lim f x, y, z  z  f x, y, z z z0 z Exercise 5.1. Find the indicated partial derivatives of the following functions by applying the definition of partial derivatives. 1. f x, y  x  y f and f x y x y 2. z  x2  y2 z and z x y 3. f x, y, z  x2  y2  z 2 f x , f y and f z 4. f x, y  1 f and f x y x y f1 and f 2 5. f x, y  1 x y References: [2] Leithold, L. (2002) [4] Peterson, T. (1964) 5.2. PARTIAL DERIVATIVES BY FORMULAS OF DIFFERENTIATION Theorems and formulas in finding the derivatives of a function presented in Chapter 2 and 3 of this module can be used in finding the derivatives of multi – valued functions. For instance, if we let f be a function defined by f x, y, z, if we look for the partial derivative of f in terms of x , then we shall hold y and z constant. Similarly, x and z will be treated constant if we look for the partial 224

MATH 401 – DIFFERENTIAL CALCULUS derivative of f in terms of y ; and finally x and y will be treated constant if we look for the partial derivative of f in terms of z . For example, if f is a function defined by f x, y, z  2x2  3y2  4z2 then the following are the partial derivatives of f : f   f x, y, z   2x2   3y2   4z 2  By applying the formulas of x x x x x Differentiation f   f x, y, z  4x  0  0 x x f   f x, y, z  4x x x f   f x, y, z   2x2   3y2   4z 2  By applying the formulas of y y y y y Differentiation f   f x, y, z  0  6y  0 y y f   f x, y, z  6y y y f   f x, y, z   2x2   3y2   4z 2  By applying the formulas of z z z z z Differentiation f   f x, y, z  0  0  8z z z f   f x, y, z  8z z z Consider the following examples: Example 5.3. Let f be a function defined by f x, y, z  2x2 y  3y2 z  4z2 x , find a. f b. f and c. f . x y z Solution: f x, y, z  2x2 y  3y2 z  4z 2 x (a.) f   2x2 y  3y2 z  4z 2 x x x x x f  2y  x2   3y2 z 4z 2  x x x x x f  2y2x  0  4z 2 1 x f  4xy  4z 2 x 225

MATH 401 – DIFFERENTIAL CALCULUS (b.) f   2x2 y  3y2 z  4z 2 x y y y y f  2x2  y  z  3y2   4z 2 x y y y y f  2x2 1  z6y  0 y f  2x2  6 yz y (c.) f   2x2 y  3y2 z  4z 2 x z z z z f   2x2 y 3y2  z  x  4z 2  z z z z f  0  3y2 1  x8z z f  3y 2  8xz z Example 5.4. Given f x, y, z  x2 y  yz 2  z3 , show that xf x  yf y  zfz  3 f x, y, z. Solution: To show that xf x  yf y  zfz  3 f x, y, z, we first find f x , f y and f z f x, y, z  x2 y  yz 2  z3 fx   x, y, z   x2 y  yz 2   z3  x x x x f x  2xy fy   x, y, z   x2 y  yz 2   z3  y y y y fy  x2  z2 fz   x, y, z   x2 y  yz 2   z3  z z z z f z  2yz  3z 2 xf x  yf y  zfz  3 f x, y, z x2xy yx2  z2  z2yz  3z2   3x2 y  yz 2  z3   2x2 y  x2 y  yz 2  2yz 2  3z3  3 x2 y  yz 2  z3  3x2 y  3yz 2  3z3  3 x2 y  yz 2  z3    3 x2 y  yz 2  z3  3 x2 y  yz 2  z3 226

MATH 401 – DIFFERENTIAL CALCULUS Example 5.5. Find f1 and f2 if f x, y  ex sin y  lnxy. Solution: f1 means partial derivative of f in terms of x or f and x f2 means partial derivative of f in terms of y or f y f1   f x, y   e x sin y  lnxy x x x f1   f x, y  sin y  ex   ln xy  x x x f1   f x, y  sin yex  y x xy f1  f  ex sin y  1 x x f2   f x, y   ex sin y  lnxy y y y f2   f x, y  ex  sin y  lnxy y y y f2   f x, y  ex sin y x y xy f2   f x, y  ex sin y  1 y y Example 5.6. Let z  coshx  2y sinh2x  y2 , find zx and z y . Solution: zx   coshx  2 y    sinh2x  y2  x x zx  sinhx  2y  x  2y  cosh2x  y2   2x  y2 x x zx  sinhx  2y1 cosh2x  y2 2  zx  sinhx  2y  2cosh2x  y2  zy   coshx  2 y    sinh2x  y2  y y zy  sinhx  2y  x  2y  cosh2x  y2   2x  y2 y y zy  sinhx  2y2 cosh2x  y2  2y  zy  2sinhx  2y  2y cosh2x  y2  227

MATH 401 – DIFFERENTIAL CALCULUS Example 5.7. If f x, y  lnx2 y  xy 2 , find f x 3,2 and f y 3,2 . Solution: For f x 3,2, we first find f x .  f x   ln x2 y  xy 2 x fx  2xy  y 2 . Substitute x3 and y2 x 2 y  xy 2 f x 3,2  232  22  12  4  8  4 32 2  322 18 12 6 3  fx 3,2  4 3 For f y 3,2, we first find f y .  f y   ln x2y  xy 2 y fy  x2  2xy . Substitute x3 and y2 x 2 y  xy 2 f y 3,2  32  232  9 12 4  2 32 2  322 18 12 6 3  f y 3,2   2 3 Example 5.8. Find Fx  0,   and Fy  0,   if Fx, y  ex tanx  y.  4   4  Solution: Fx  ex   tanx  y tanx  y  ex x x Fx  ex sec2 x  y  ex tanx  y Fx  0,    e 0 sec 2  0     e 0 t an 0     4   4   4  Fx  0,    1sec2     1 tan    4   4   4   Fx  0,    2  4  2 1  Fx  0,    1  4  Fy  ex   tanx  y  tanx  y  ex y y Fy  ex  sec2 x  y  tanx  y0 Fy  ex sec2 x  y 228

MATH 401 – DIFFERENTIAL CALCULUS Fy  0,    e 0 sec 2  0     4   4   Fy 0,    1 2  4  2  Fy  0,    2  4  Example 5.9. Find the slopes of the surface given by f x, y  1 x 12  y 12 at the point 1,2 in the x  direction and y  direction. Solution: We first find the partial derivatives f x and f y . That is f x, y  1 x 12  y 12 f x  2x 1 at the point 1,2 ; fx 1,2  211  0 Therefore, the slope of the surface at the point 1,2 on the x  direction is 0. f x, y  1 x 12  y 12 f y  2y 1 at the point 1,2 ; fx 1,2  22 1  2 Therefore, the slope of the surface at the point 1,2 on the y  direction is  2 . Exercise 5.2. Find the first partial derivatives of the following. 1. f x, y  x  y x y 2. z  x2  y 2 3. f x, y, z  x2  y2  z 2 4. f x, y  1 x y 5. f x, y  1 x y  6. z  ln x  x2  y2 7. f x, y  Sin 1 1 x2 y2 8. If z  x2 y 2 , show that x z  y z  3z . x y x y 229

MATH 401 – DIFFERENTIAL CALCULUS 9. If u  exy , show that u  u  u . ex  ey x y 10.If u  x2 y  y2 z  z 2 x , show that ux  uy  uz  x  y  z2 . 5.3. HIGHER – ORDER PARTIAL DERIVATIVES Since the partial derivatives of a function f x, y are expressed in terms of x and y , they can still be differentiated with respect to the same variables. For instance the partial derivatives f or f can still be differentiated with respect to x and x y y i.e.)   f  is the second partial derivative of f in terms of x ;   f  is the x  x  y  x  partial derivative of f with respect to y , so that is the second partial x derivatives of f . Let f be a multi – valued function defined by f x, y, z, the following are the first partial derivatives of f : fx  f ; fy  f and fz  f x y z The second partial derivatives of f are the partial derivatives of f x , f y and f z . So we have   f   2 f  f xx   f   2f  f xy   f    2 f  f xz x  x  x y  x  yx z  x  zx 2   f   2f  f yx   f   2 f  f yy   f   2f  f yz x y xy y y y z y zy 2   f    2 f  f zx   f   2f  f zy   f   2 f  f zz x  z  xz y  z  yz z  z  z 2 Example 5.10. Find the second partial derivatives of f in Example 5.3. Solution: f in Example 5.3 is defined by f x, y, z  2x2 y  3y2 z  4z2 x and f  4xy  4z 2 ; f  2x2  6 yz and f  3y 2  8xz . So the second partial x y z derivatives of f are:   f   2 f  f xx  4 y   f   2f  f xy  4x   f   2f  f xz  8z x  x  x y  x  yx z  x  zx 2   f   2f  f yx  4x   f   2f  f yy  6z   f   2f  f yz  6 y x y xy y y y2 z y zy 230

MATH 401 – DIFFERENTIAL CALCULUS   f    2 f  f zx  8z   f   2f  f zy  6 y   f   2 f  f zz  8x x  z  xz y  z  yz z  z  z 2 Example 5.11. Find f12 , f21, f11 and f22 if f x, y  ex sin y  lnxy. Solution: From Example 5.5, we see that f1  f  ex sin y  1 and f2   f x, y  ex sin y  1 x x y y f12    f   2 f and f 21    f   2 f . Then we have y  x  yx x y xy f12    e x sin y  1   ex cos y y  x  f 21   e x sin y  1   ex sin y x y f11    f   2 f and f 22    f   2 f x  x  x 2 y y y 2 f11    e x sin y  1   ex sin y  1 x  x  x2 f 22   e x sin y  1   ex cos y  1 y y y2 Example 5.12. If z  lnx2  y2 , show that 2 z  2 z  0 . x 2 y 2 Solution: If z  lnx2  y2 , then the first partial derivatives of z are z  2x and z  2y . x x 2  y 2 y x2  y 2 Differentiating these with respect to x and y respectively, we get the second partial derivatives as follows:  2 z  x2  y 2 2  2x2x  2x2  2 y 2  4x2  2 y 2  2x2      x2 x2  y2 2 x2  y2 2 x2  y2 2  2 z  x2  y 2 2  2 y2 y  2x2  2 y 2  4y 2  2x2  2 y 2      y2 x2  y2 2 x2  y2 2 x2  y2 2 We now show that 2 z   2 z  0 x 2 y 2    2 y 2  2x2  2x2  2 y 2  0 x2  y2 2 x2  y2 2 00 231

MATH 401 – DIFFERENTIAL CALCULUS Example 5.13. If z  Sin 1 y  , show that  2 z  2 z .  x  xy yx Solution: If z  Sin 1 y  , then the first partial derivatives of z are x z  y y y y y (1) x2  x2  x2  x2  (2) x 1   y 2 1 y2 x2  y2 x2  y2 x x2  y2 x x2 x2 x 11 1 1 x x z  x  x  x2  y2 x2  y2 1 y 1   y 2 1 y2 x2 x x2  y2 x x2 Differentiating (1) with respect to y and (2) with respect to x , we get the second partial derivatives of z .    2z 1 x x2  y2 1   y 1 x x2  y2  2  2 y   2   yx 2 x x2  y2   x  2 z       yx x 2  y 2  xy 2  x x 2  y 2  xy 2 x2  y2  x2  y2   x x2  y2  y2 2 23 x x2  y2 x x2  y2 x2 x2  y2  2 z  x (3) yx 3 x2  y2    2z   2x  xy 3 x (4) 2 x2  y2 3 x2  y2 It can be seen that (3) and (4) are equal, therefore 2 z  2 z . xy yx Exercise 5.3. Find all the second partial derivatives of the following functions. 1. z  3xy 2 2. f x, y  x2  2xy  3y2 3. z  x2  y 2 4. f x, y, z  ex tan y cos z 232

MATH 401 – DIFFERENTIAL CALCULUS 5. f x, y  2xe y  3yex 6. If z  x  y , show that  2 z  2 z . x  y xy yx  7. If z  ln x  x2  y2 , show that 2 z  2 z . xy yx 8. If z  ex sin y  e y sin x , show that 2 z   2 z  0 . x 2 y 2 9. If z  Tan1 y  , show that 2z  2z 0.  x 2 y 2 x  10. If u  x2  y2 3 , verify that 3u  3u . x2y yx2 References: [2] Leithold, L. (2002) [4] Peterson, T. (1964) 5.4. TOTAL DERIVATIVES Let z be a function in terms of x and y , z  f x, y, and let x and y be the arbitrary increments of x and y respectively, then we can write an increment in z as z  f x  x, y  y f x, y (1) Also, we can find the total derivatives of z by applying the following definition. Definition 5.3. [1] If z  f x, y and x and y are the increments of x and y , then the differentials of the independent variables x and y are dx  x and dy  y and the total derivatives of the independent variable z is given by dz  z dx  z dy  f x x, ydx  f y x, ydy . (2) x y This definition can be extended into functions of more than two variables, say w  f x, y, z. The total derivative of w , denoted by dw is given by dw  w dx  w dy  w dz (3) x y z Example 5.14. Find the total derivative of z  x2e2y  ye x . Solution: Applying Definition 5.3, we have dz  z dx  z dy x y    dz  2xe2y  ye x dx  2x2e2y  ex dy 233

MATH 401 – DIFFERENTIAL CALCULUS Example 5.15. Find the total derivative of w  3x2  y2  2z2 . Solution: Following (2), the total derivative of w is dw  w dx  w dy  w dz , so we x y z have dw  6xdx  2ydy  4zdz Definition 5.3 can be extended into a case where x and y are continuous functions express in terms of the third variable, say t ; that is x  f t and y  gt (4) If the values of (4) is substituted to z  f x, y, z becomes a function expressed in terms of t . To find the total derivative of z in terms of t , we shall only divide (2) by the differential of t , that is dt , thus dz  z  dx  z  dy (5) dt x dt y dt Similarly if x, y and z in (3) are continuous functions in terms of t , then the total derivative of w is dw  w  dx  w  dy  w  dz (6) dt x dt y dt z dt Example 5.16. If z  x ln y ; x  2  u2 and y  eu find dz . du Solution: By (5) dz  z  dx  z  dy . We have z  ln y , z  x ; dx  2u and dy  eu du x du y du x y y du du dz  ln y 2u    x eu  du y dz  ln eu 2u    2  u2 eu  Simplifying du eu dz  2u 2  2  u 2 du  dz  3u 2  2 du 234

MATH 401 – DIFFERENTIAL CALCULUS Example 5.17. [4] The height of a right circular cylinder is 50 inches and decreases at the rate of 4 inches per second, while the radius of the bases is 20 inches and increases at the rate of 1 inch per second. At what rate is the volume changing? Solution: Let V  r 2h where r is the radius and h is the height. Differentiating V partially in terms of the time t , we have dV  2rh dr  r 2 dh dt dt dt Since h  50 , r  20 ; dh  4 and dr  1 , then dt dt dV  220501  202  4 dt dV  400 . dt Thus, the volume is increasing at a rate of 400 cubic inches per second. Example 5.18. Find du if u  Tan1xy; x  sin t and y  sect dt Solution: du  u  dx  u  dy dt x dt y dt du   1  y cos t  1 x sec t tan t  dt x2 y2 x2 y2 du   sec t t cos t   sin t t sec t tan t  dt 1 sin 2 t sec2 1 sin 2 t sec2   du  1 1 t   1 sin t 2 t sec t tan t  dt t an 2   tan  du  1 t   sin t  sin t  dt sec2  sect  cos t  du  cos 2 t  sin 2 t dt du  1 dt Example 5.19. Find the total derivative of z if z  x2  2xy  y2 ; x  t 12 and y  t 12 Solution: dz  z  dx  z  dy dt x dt y dt 235

MATH 401 – DIFFERENTIAL CALCULUS dz  2x  2y2t 1  2x  2y2t 1 dt  dz  22x  2yt 1 t 1  4 t 12  t 12 2 dt dz  8t 2  2t 1 t 2  2t 1  84t dt  dz  32t dt Exercise 5.4. Find du in each of the following. dt 1. u  Tan1 y  x  ln t y  et z t x x  t cos t y  t sin t 2. u  xy  xz  yz 3. u  x  t x  ln t y  ln1 yt t Find du in each of the following dx 4. u  x  y y  x2 1 y  5. u  ln x2  y2  z2 y  xsin x z  xcos x References: [2] Leithold, L. (2002) [4] Peterson, T. (1964) 5.5. CHAIN RULE OF PARTIAL DIFFERENTIATIONS Suppose x and y in (4) are expressed in two variables, say r and t , i.e. x  f r,t and y  gr,t, then we can now express z as a function of r and t or z  f r,t. By applying the chain rule of partial differentiation we get z  z  x  z  y t x t y t z  z  x  z  y r x r y r Similarly if x, y and z in (3) are continuous functions in terms of r and t , then the partial derivative of w is given by w  w  x  w  y  w  z t x t y t z t 236

MATH 401 – DIFFERENTIAL CALCULUS w  w  x  w  y  w  z r x r y r z r Definition 5.4. [2] Suppose that u is a differentiable function of the n variables x1, x2 ,...xn1, xn and each of these variables is in turn a function of the m variables y1, y2 ,...ym1, ym . Suppose further that each of the partial derivatives xi where i  1,2,3,..., n 1, n and j  1,2,3,..., m 1, m ; exists. Then u is a function x j of y1, y2 ,...ym1, ym and u  u x1  u x2  ...  u xn y1 x1 y1 x2 y1 xn y1 u  u x1  u x2  ...  u xn y2 x1 y2 x2 y2 xn y2  u  u x1  u x2  ...  u xn ym x1 ym x2 ym xn ym Example 5.20. Let u  x2  yz ; x  r sin t ; y  r cos t and z  r sin2 t find u and u . r t Solution: By Definition 5.4, we have u  u  x  u  y  u  z and u  u  x  u  y  u  z r x r y r z r t x t y t z t u  2xsin t zcos t ysin2 t r u  2r sin tsin t  r sin 2 tcos t  r cos tsin2 t r u  2r sin 2 t  2r sin 2 t cos t r u  2r sin 2 t1 cos t r u  2xr cos t  z r sin t  y2r sin t cos t t u  2r sin tr cos t  r sin 2 t r sin t  r cos t2r sin t cos t t u  2r 2 sin t cos t  r 2 sin3 t  2r 2 sin t cos 2 t t 237

MATH 401 – DIFFERENTIAL CALCULUS  u  r 2 sin t 2cos t  sin 2 t  2cos 2 t t Example 5.21. Given y x  2r cos t and y  4r sin t , find u and u . uex ; r t Solution: u  u  x  u  y r x r y r u    y y 2 cos t   1 y 4 sin t r x2 x ex ex u    4r sin t t 4r sin t 2 cos t   2r 1 4r sin t 4 sin t r 4r 2 cos 2 cos t e 2r cost e 2r cost u   2 sin t e 2 tan t  2 sin t e 2 tan t r r cos t r cos t u  0 r u  u  x  u  y t x t y t u    y e y  2r sin t   1 y 4r cos t t x2 x x ex u    4r sin t t 4r sin t  2r sin t  2r 1 4r sin t 4r cos t t 4r 2 cos 2 cos t e 2r cost e 2r cost u  2 sin 2 t e 2 tan t  2e 2 tan t t cos 2 t u  2 tan 2 te 2 tan t  2e 2 tan t t  u  2e2 tan t tan 2 t  1 t Example 5.22. Let u  r 2  s2 ; r  x  y and s  x  y , find u and u . x y Solution: u  u  r  u  s u  u  r  u  s x r x s x y r y s y u  2r1  2s1 u  2r1  2s1 x y u  2r  s u  2r  s x y u  2x  y  x  y u  2x  y  x  y x y 238

MATH 401 – DIFFERENTIAL CALCULUS u  4x u  4 y x x Implicit Partial Derivatives If z  f x, y and y is a function of x , it follows from Section 5.4 that dz  f  f  dy dx x y dx For z  0 , identically, we have dz  0 ; hence dx 0  f  f  dy x y dx Solving for dy , we have dx f dy   x where f  0 . dx f y y Example 5.23. If x sin y  y cos x  3 , find dy . dx Solution: If we first let f x, y  x sin y  y cos x  3  0 . SO we have f dy  x   sin y  y sin x dx f x cos cos x y If z is defined as an implicit function of x and y by the equation Fx, y, z  0 , then z   Fx z   Fy and y   Fx where Fz 0 x Fz y Fz x Fy Example 5.24. If x3  y3  z3  3xyz  2 , find z , z and y . x y x Solution: We first write x3  y3  z 3  3xyz  2 as Fx, y, z  x3  y3  z3  3xyz  2  0 239

MATH 401 – DIFFERENTIAL CALCULUS z   Fx   3x2  3yz   x2  yz x Fz 3z 2  3xy z 2  xy z   Fy   3y 2  3xz   y 2  xz y Fz 3z 2  3xy z 2  xy y   Fx   3x2  3yz   x2  yz x Fy 3y 2  3xz y 2  xz Example 5.25. Let x y y x 1, find y . x Solution: y   Fx x Fy We first take the logarithm of both side of the equation. lnx y y x   ln1 Applying the properties of logarithmic function lnx y   lny x   0 y ln x  x ln y  0 We shall now let F  y ln x  x ln y  0 y   Fx y  ln y  y  x ln y  yy  x ln y x Fy  x xy ln x  x x ln x  x y ln x  x yy Exercise 5.5. Find the indicated partial derivatives of the following functions by applying the chain rule of partial differentiation. 1. u  p  qr ; p  x2 ; q  xy ; r  y 2 ; u and u x y 2. w  xyz ; x  s  t ; y  s  t ; z  st 2 ; w and w s t 3. w  x cos yz ; x  s 2 ; y  t 2 ; z  s  2t ; w and w s t 4. w  x2  y 2  z 2 ; x  t sin r ; y  t cos r ; z  rt 2 w and w r t 5. u  Tan1 y  ; x  r cos  ; y  r sin  ; u and u  x r  6. ye 2x  xe 2y  0 ; dy dx 7. x sin y  y cos z  3z sin x  0 ; z , z and y x y x 8.  ln x2  y 2  ez 1; z , z and y x y x 240

MATH 401 – DIFFERENTIAL CALCULUS 9. x  y  1 ; z , z and y x y x x y z 10. 4x2  2xy  y 2 1; y x References: [2] Leithold, L. (2002) [4] Peterson, T. (1964) CHAPTER TEST Find the indicated partial derivatives of the following functions by applying the definition of partial derivatives. 1. f x, y  x2  2xy  3y 2 f and f 2. z  x  y x y 3. f x, y, z  x2  y 2  z 2  2xz z and z x y f x , f y and f z Find the first partial derivatives of the following by applying the formulas of differentiation. 4. f x, y  x2  2xy  3y2 5. f x, y  xe y  ye 2x 6. z  1 y2  lnx2  xy z 7. If u  e y  e z  e x , show that xux  yu y  zuz  0 . 8. If z  ey cosx  y , show that zx  z y  z  0 9. If z  e y cos x  ex cos y , find  2 z and  2 z x 2 y 2 10. If z  Sin 1 x  , find 2z and 2z . y x 2 y 2  11. If u  x3  y3 2 , find 3u and 3u x 2y yx 2 Find the indicated derivatives 12. ex  e y  ez  exyz ; z , z and y x y x 241

MATH 401 – DIFFERENTIAL CALCULUS 13. z  yTan1xz; z , z and y x y x 14. If u  Tan1 x  ; x  et ; y  ln t find du . y dt 15. If u  xy  xz  yz ; x  t sin t ; y  t cos t and z  t , find du . dt 16.If w  x3  y3  z3 ; x  t sin r ; y  t cos r ; and z  rt 2 , find w and w . r t 17.If w  x2 y2 z 2 ; x  s  t ; y  s  t ; and z  st ; find w and w . s t 18. Find du if u  x  y and y  2x2 dx 1 y  19. Find du if u  ln x2  y2  z2 ; y  xsin x ; and z  xcos x . dx  20. Find du if u  ln x2  y2  z2 ; y  xsin x ; and z  xcos x . dz References: [2] Leithold, L. (2002) [4] Peterson, T. (1964) 242

MATH 401 – DIFFERENTIAL CALCULUS References [1] Larson, R. (2018), Calculus. 11th Edition. Cengage Learning Asia Pte. Ltd. [2] Leithold, Louis (2002). The Calculus 7. Pearson Education Asia Pte. Ltd [3] Stewart, James. (2016). Calculus: Early Transcendentals. 8th Edition. Cengage Learning [4] http://www.copingwithcalculus.com/DifferentiationApplications.html [5] https://themathpage.com/aCalc/applied.htm [6] https://mathalino.com/reviewer/differential-calculus [7] https://www.intmath.com/differentiation-transcendental/4- applications-derivatives-trigonometric.php 243