1.E: Understanding the Derivative (Exercises) - Mathematics

1.1: How do we Measure Velocity?

1. A bungee jumper dives from a tower at time t = 0. Her height h (measured in feet) at time t (in seconds) is given by the graph in Figure 1.3. Figure 1.3: A bungee jumper’s height function

In this problem, you may base your answers on estimates from the graph or use the fact that the jumper’s height function is given by s(t) = 100 cos(0.75t) · e −0.2t + 100.

(a) What is the change in vertical position of the bungee jumper between t = 0 and t = 15?

(b) Estimate the jumper’s average velocity on each of the following time intervals: [0, 15], [0, 2], [1, 6], and [8, 10]. Include units on your answers.

(c) On what time interval(s) do you think the bungee jumper achieves her greatest average velocity? Why?

(d) Estimate the jumper’s instantaneous velocity at t = 5. Show your work and explain your reasoning, and include units on your answer.

(e) Among the average and instantaneous velocities you computed in earlier questions, which are positive and which are negative? What does negative velocity indicate?

2. A diver leaps from a 3 meter springboard. His feet leave the board at time t = 0, he reaches his maximum height of 4.5 m at t = 1.1 seconds, and enters the water at t = 2.45. Once in the water, the diver coasts to the bottom of the pool (depth 3.5 m), touches bottom at t = 7, rests for one second, and then pushes off the bottom. From there he coasts to the surface, and takes his first breath at t = 13.

(a) Let s(t) denote the function that gives the height of the diver’s feet (in meters) above the water at time t. (Note that the “height” of the bottom of the pool is −3.5 meters.) Sketch a carefully labeled graph of s(t) on the provided axes in Figure 1.4. Include scale and units on the vertical axis. Be as detailed as possible. Figure 1.4: Axes for plotting s(t) in part (a) and v(t) in part (c) of the diver problem.

(b) Based on your graph in (a), what is the average velocity of the diver between t = 2.45 and t = 7? Is his average velocity the same on every time interval within [2.45, 7]?

(c) Let the function v(t) represent the instantaneous vertical velocity of the diver at time t (i.e. the speed at which the height function s(t) is changing; note that velocity in the upward direction is positive, while the velocity of a falling object is negative). Based on your understanding of the diver’s behavior, as well as your graph of the position function, sketch a carefully labeled graph of v(t) on the axes provided in Figure 1.4. Write several sentences that explain how you constructed your graph, discussing when you expect v(t) to be zero, positive, negative, relatively large, and relatively small.

(d) Is there a connection between the two graphs that you can describe? What can you say about the velocity graph when the height function is increasing? decreasing? Make as many observations as you can.

3. According to the U.S. census, the population of the city of Grand Rapids, MI, was 181,843 in 1980; 189,126 in 1990; and 197,800 in 2000.

(a) Between 1980 and 2000, by how many people did the population of Grand Rapids grow?

(b) In an average year between 1980 and 2000, by how many people did the population of Grand Rapids grow?

(c) Just like we can find the average velocity of a moving body by computing change in position over change in time, we can compute the average rate of change of any function f . In particular, the average rate of change of a function f over an interval [a, b] is the quotient f (b) − f (a) b − a . What does the quantity f (b)−f (a) b−a measure on the graph of y = f (x) over the interval [a, b]?

(d) Let P(t) represent the population of Grand Rapids at time t, where t is measured in years from January 1, 1980. What is the average rate of change of P on the interval t = 0 to t = 20? What are the units on this quantity?

(e) If we assume the population of Grand Rapids is growing at a rate of approximately 4% per decade, we can model the population function with the 10 formula P(t) = 181843(1.04) t/10 . Use this formula to compute the average rate of change of the population on the intervals [5, 10], [5, 9], [5, 8], [5, 7], and [5, 6].

(f) How fast do you think the population of Grand Rapids was changing on January 1, 1985? Said differently, at what rate do you think people were being added to the population of Grand Rapids as of January 1, 1985? How many additional people should the city have expected in the following year? Why?

1.2: The Notion of Limit

1. Consider the function whose formula is f (x) = 16 − x 4 x 2 − 4 . (a) What is the domain of f? (b) Use a sequence of values of x near a = 2 to estimate the value of lim x→2 f (x), if you think the limit exists. If you think the limit doesn’t exist, explain why. (c) Evaluate limx→2 f (x) exactly, if the limit exists, or explain how your work shows the limit fails to exist. Here you should use algebra to factor and simplify the numerator and denominator of f (x) as you work to evaluate the limit. Discuss how your findings compare to your results in (b). (d) True or false: f (2) = −8. Why? (e) True or false: 16−x 4 x 2−4 = −4 − x 2 . Why? How is this equality connected to your work above with the function f? (f) Based on all of your work above, construct an accurate, labeled graph of y = f (x) on the interval [1, 3], and write a sentence that explains what you now know about lim x→2 16 − x 4 x 2 − 4 .

2. Let g(x) = − |x + 3| x + 3 . (a) What is the domain of g? (b) Use a sequence of values near a = −3 to estimate the value of limx→−3 g(x), if you think the limit exists. 20 (c) Evaluate limx→2 g(x) exactly, if the limit exists, or explain how your work shows the limit fails to exist. Here you should use the definition of the absolute value function in the numerator of g(x) as you work to evaluate the limit. (Hint: |a| = a whenever a ≥ 0, but |a| = −a whenever a < 0.) (d) True or false: g(−3) = −1. Why? (e) True or false: − |x+3| x+3 = −1. Why? How is this equality connected to your work above with the function g? (f) Based on all of your work above, construct an accurate, labeled graph of y = g(x) on the interval [−4, −2], and write a sentence that explains what you now know about lim x→−3 g(x).

3. For each of the following prompts, sketch a graph on the provided axes of a function that has the stated properties. Figure 1.9: Axes for plotting y = f (x) in (a) and y = g(x) in (b).

(a) y = f (x) such that

• f (−2) = 2 and lim x→−2 f (x) = 1

• f (−1) = 3 and lim x→−1 f (x) = 3

• f (1) is not defined and lim x→1 f (x) = 0

• f (2) = 1 and lim x→2 f (x) does not exist. (b) y = g(x) such that • g(−2) = 3, g(−1) = −1, g(1) = −2, and g(2) = 3

• At x = −2, −1, 1 and 2, g has a limit, and its limit equals the value of the function at that point. 21

• g(0) is not defined and lim x→0 g(x) does not exist.

4. Her height s in feet at time t in seconds is given by s(t) = 100 cos(0.75t) · e −0.2t + 100. (a) Write an expression for the average velocity of the bungee jumper on the interval [1, 1 + h]. (b) Use computing technology to estimate the value of the limit as h → 0 of the quantity you found in (a). (c) What is the meaning of the value of the limit in (b)? What are its units?

1.3: The Derivative of a Function at a Point

1. Consider the graph of y = f (x) provided in Figure 1.16. (a) On the graph of y = f (x), sketch and label the following quantities: 31

• the secant line to y = f (x) on the interval [−3, −1] and the secant line to y = f (x) on the interval [0, 2].

• the tangent line to y = f (x) at x = −3 and the tangent line to y = f (x) at x = 0. Figure 1.16: Plot of y = f (x).

(b) What is the approximate value of the average rate of change of f on [−3, −1]? On [0, 2]? How are these values related to your work in (a)?

(c) What is the approximate value of the instantaneous rate of change of f at x = −3? At x = 0? How are these values related to your work in (a)? 2. For each of the following prompts, sketch a graph on the provided axes in Figure 1.17 of a function that has the stated properties. (a) y = f (x) such that • the average rate of change of f on [−3, 0] is −2 and the average rate of change of f on [1, 3] is 0.5, and

• the instantaneous rate of change of f at x = −1 is −1 and the instantaneous rate of change of f at x = 2 is 1. (b) y = g(x) such that

• g(3)−g(−2) 5 = 0 and g(1)−g(−1) 2 = −1, and

• g 0 (2) = 1 and g 0 (−1) = 0

3. Suppose that the population, P, of China (in billions) can be approximated by the function P(t) = 1.15(1.014) t where t is the number of years since the start of 1993.

(a) According to the model, what was the total change in the population of China between January 1, 1993 and January 1, 2000? What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January 1, 2000? Explain and justify, being sure to include proper units on all your answers. Figure 1.17: Axes for plotting y = f (x) in (a) and y = g(x) in (b).

(b) According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, 2012?

(c) Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today’s date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found.

(d) Find an equation for the tangent line to the function y = P(t) at the point where the t-value is given by today’s date.

4. The goal of this problem is to compute the value of the derivative at a point for several different functions, where for each one we do so in three different ways, and then to compare the results to see that each produces the same value. For each of the following functions, use the limit definition of the derivative to compute the value of f 0 (a) using three different approaches: strive to use the algebraic approach first (to compute the limit exactly), then test your result using numerical evidence (with small values of h), and finally plot the graph of y = f (x) near (a, f (a)) along with the appropriate tangent line to estimate the value of f 0 (a) visually. Compare your findings among all three approaches; if you are unable to complete the algebraic approach, still work numerically and graphically.

(a) f (x) = x 2 − 3x, a = 2

(b) f (x) = 1 x , a = 1

(c) f (x) = √ x, a = 1 33

(d) f (x) = 2 − |x − 1|, a = 1

(e) f (x) = sin(x), a = π 2

1.4: The Derivative Function

1. Let f be a function with the following properties: f is differentiable at every value of x (that is, f has a derivative at every point), f (−2) = 1, and f 0 (−2) = −2, f 0 (−1) = −1, f 0 (0) = 0, f 0 (1) = 1, and f 0 (2) = 2.

(a) On the axes provided at left in Figure 1.19, sketch a possible graph of y = f (x). Explain why your graph meets the stated criteria.

(b) On the axes at right in Figure 1.19, sketch a possible graph of y = f 0 (x). What type of curve does the provided data suggest for the graph of y = f 0 (x)?

(c) Conjecture a formula for the function y = f (x). Use the limit definition of the derivative to determine the corresponding formula for y = f 0 (x). Discuss both graphical and algebraic evidence for whether or not your conjecture is correct. Figure 1.19: Axes for plotting y = f (x) in (a) and y = f 0 (x) in (b).

2. Consider the function g(x) = x 2 − x + 3.

(a) Use the limit definition of the derivative to determine a formula for g 0 (x). 41

(b) Use a graphing utility to plot both y = g(x) and your result for y = g 0 (x); does your formula for g 0 (x) generate the graph you expected?

(c) Use the limit definition of the derivative to find a formula for p 0 (x) where p(x) = 5x 2 − 4x + 12.

(d) Compare and contrast the formulas for g 0 (x) and p 0 (x) you have found. How do the constants 5, 4, 12, and 3 affect the results?

3. Let g be a continuous function (that is, one with no jumps or holes in the graph) and suppose that a graph of y = g 0 (x) is given by the graph on the right in Figure 1.20. Figure 1.20: Axes for plotting y = g(x) and, at right, the graph of y = g 0 (x).

(a) Observe that for every value of x that satisfies 0 < x < 2, the value of g 0 (x) is constant. What does this tell you about the behavior of the graph of y = g(x) on this interval?

(b) On what intervals other than 0 < x < 2 do you expect y = g(x) to be a linear function? Why?

(c) At which values of x is g 0 (x) not defined? What behavior does this lead you to expect to see in the graph of y = g(x)? (d) Suppose that g(0) = 1. On the axes provided at left in Figure 1.20, sketch an accurate graph of y = g(x). 42 4. For each graph that provides an original function y = f (x) in Figure 1.21 (on the following page), your task is to sketch an approximate graph of its derivative function, y = f 0 (x), on the axes immediately below. View the scale of the grid for the graph of f as being 1 × 1, and assume the horizontal scale of the grid for the graph of f 0 is identical to that for f . If you need to adjust the vertical scale on the axes for the graph of f 0 , you should label that accordingly. Figure 1.21: Graphs of y = f (x) and grids for plotting the corresponding graph of y = f 0 (x).

1.5: Interpretating, Estimating, and Using the Derivative

1. A cup of coffee has its temperature F (in degrees Fahrenheit) at time t given by the function F(t) = 75 + 110e −0.05t , where time is measured in minutes.

(a) Use a central difference with h = 0.01 to estimate the value of F 0 (10).

(b) What are the units on the value of F 0 (10) that you computed in (a)? What is the practical meaning of the value of F 0 (10)?

(c) Which do you expect to be greater: F 0 (10) or F 0 (20)? Why?

(d) Write a sentence that describes the behavior of the function y = F 0 (t) on the time interval 0 ≤ t ≤ 30. How do you think its graph will look? Why?

2. The temperature change T (in Fahrenheit degrees), in a patient, that is generated by a dose q (in milliliters), of a drug, is given by the function T = f (q).

(a) What does it mean to say f (50) = 0.75? Write a complete sentence to explain, using correct units.

(b) A person’s sensitivity, s, to the drug is defined by the function s(q) = f 0 (q). What are the units of sensitivity?

(c) Suppose that f 0 (50) = −0.02. Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).

3. The velocity of a ball that has been tossed vertically in the air is given by v(t) = 16−32t, where v is measured in feet per second, and t is measured in seconds. The ball is in the air from t = 0 until t = 2.

(a) When is the ball’s velocity greatest?

(b) Determine the value of v 0 (1). Justify your thinking.

(c) What are the units on the value of v 0 (1)? What does this value and the corresponding units tell you about the behavior of the ball at time t = 1?

(d) What is the physical meaning of the function v 0 (t)?

4. The value, V, of a particular automobile (in dollars) depends on the number of miles, m, the car has been driven, according to the function V = h(m).

(a) Suppose that h(40000) = 15500 and h(55000) = 13200. What is the average rate of change of h on the interval [40000, 55000], and what are the units on this value? 51

(b) In addition to the information given in (a), say that h(70000) = 11100. Determine the best possible estimate of h 0 (55000) and write one sentence to explain the meaning of your result, including units on your answer.

(c) Which value do you expect to be greater: h 0 (30000) or h 0 (80000)? Why?

(d) Write a sentence to describe the long-term behavior of the function V = h(m), plus another sentence to describe the long-term behavior of h 0 (m). Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.

1.6: The Second Derivative

1. Suppose that y = f (x) is a differentiable function for which the following information is known: f (2) = −3, f 0 (2) = 1.5, f 00(2) = −0.25.

(a) Is f increasing or decreasing at x = 2? Is f concave up or concave down at x = 2?

(b) Do you expect f (2.1) to be greater than −3, equal to −3, or less than −3? Why?

(c) Do you expect f 0 (2.1) to be greater than 1.5, equal to 1.5, or less than 1.5? Why?

(d) Sketch a graph of y = f (x) near (2, f (2)) and include a graph of the tangent line. 63

2. For a certain function y = g(x), its derivative is given by the function pictured in Figure 1.34. Figure 1.34: The graph of y = g 0 (x).

(a) What is the approximate slope of the tangent line to y = g(x) at the point (2, g(2))?

(b) How many real number solutions can there be to the equation g(x) = 0? Justify your conclusion fully and carefully by explaining what you know about how the graph of g must behave based on the given graph of g 0 .

(c) On the interval −3 < x < 3, how many times does the concavity of g change? Why?

(d) Use the provided graph to estimate the value of g 00(2). 3. A bungee jumper’s height h (in feet ) at time t (in seconds) is given in part by the data in the following table:

t 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 h(t) 200 184.2 159.9 131.9 104.7 81.8 65.5 56.8 55.5 60.4 69.8 t 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 9.5 10.0 h(t) 81.6 93.7 104.4 112.6 117.7 119.4 118.2 114.8 110.0 104.7

(a) Use the given data to estimate h 0 (4.5), h 0 (5), and h 0 (5.5). At which of these times is the bungee jumper rising most rapidly?

(b) Use the given data and your work in (a) to estimate h 00(5).

(c) What physical property of the bungee jumper does the value of h 00(5) measure? What are its units?

(d) Based on the data, on what approximate time intervals is the function y = h(t) concave down? What is happening to the velocity of the bungee jumper on these time intervals? 64

4. For each prompt that follows, sketch a possible graph of a function on the interval −3 < x < 3 that satisfies the stated properties.

(a) y = f (x) such that f is increasing on −3 < x < 3, f is concave up on −3 < x < 0, and f is concave down on 0 < x < 3.

(b) y = g(x) such that g is increasing on −3 < x < 3, g is concave down on −3 < x < 0, and g is concave up on 0 < x < 3.

(c) y = h(x) such that h is decreasing on −3 < x < 3, h is concave up on −3 < x < −1, neither concave up nor concave down on −1 < x < 1, and h is concave down on 1 < x < 3.

(d) y = p(x) such that p is decreasing and concave down on −3 < x < 0 and p is increasing and concave down on 0 < x < 3.

1.7: Limits, Continuity, and Differentiability

1. Consider the graph of the function y = p(x) that is provided in Figure 1.42. Assume that each portion of the graph of p is a straight line, as pictured. Figure 1.42: At left, the piecewise linear function y = p(x). At right, axes for plotting y = p 0 (x).

(a) State all values of a for which limx→a p(x) does not exist.

(b) State all values of a for which p is not continuous at a.

(c) State all values of a for which p is not differentiable at x = a.

(d) On the axes provided in Figure 1.42, sketch an accurate graph of y = p 0 (x).

2. For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why. 75

(a) A function f that is continuous at a = 2 but not differentiable at a = 2.

(b) A function g that is differentiable at a = 3 but does not have a limit at a = 3.

(c) A function h that has a limit at a = −2, is defined at a = −2, but is not continuous at a = −2.

(d) A function p that satisfies all of the following:

• p(−1) = 3 and limx→−1 p(x) = 2

• p(0) = 1 and p 0 (0) = 0

• limx→1 p(x) = p(1) and p 0 (1) does not exist

3. Let h(x) be a function whose derivative y = h 0 (x) is given by the graph on the right in Figure 1.43.

(a) Based on the graph of y = h 0 (x), what can you say about the behavior of the function y = h(x)?

(b) At which values of x is y = h 0 (x) not defined? What behavior does this lead you to expect to see in the graph of y = h(x)?

(c) Is it possible for y = h(x) to have points where h is not continuous? Explain your answer.

(d) On the axes provided at left, sketch at least two distinct graphs that are possible functions y = h(x) that each have a derivative y = h 0 (x) that matches the provided graph at right. Explain why there are multiple possibilities for y = h(x). Figure 1.43: Axes for plotting y = h(x) and, at right, the graph of y = h 0 (x).

4. Consider the function g(x) = p |x|.

(a) Use a graph to explain visually why g is not differentiable at x = 0. 76

(b) Use the limit definition of the derivative to show that g 0 (0) = lim h→0 p |h| h .

(c) Investigate the value of g 0 (0) by estimating the limit in (b) using small positive and negative values of h. For instance, you might compute √ |−0.01| 0.01 . Be sure to use several different values of h (both positive and negative), including ones closer to 0 than 0.01. What do your results tell you about g 0 (0)?

(d) Use your graph in (a) to sketch an approximate graph of y = g 0 (x).

1.8: The Tangent Line Approximation

1. A certain function y = p(x) has its local linearization at a = 3 given by L(x) = −2x + 5. 84

(a) What are the values of p(3) and p 0 (3)? Why?

(b) Estimate the value of p(2.79).

(c) Suppose that p 00(3) = 0 and you know that p 00(x) < 0 for x < 3. Is your estimate in (b) too large or too small?

(d) Suppose that p 00(x) > 0 for x > 3. Use this fact and the additional information above to sketch an accurate graph of y = p(x) near x = 3. Include a sketch of y = L(x) in your work.

2. A potato is placed in an oven, and the potato’s temperature F (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time t is measured in minutes.

t F(t) 0 70 15 180.5 30 251 45 296 60 324.5 75 342.8 90 354.5

(a) Use a central difference to estimate F 0 (60). Use this estimate as needed in subsequent questions.

(b) Find the local linearization y = L(t) to the function y = F(t) at the point where a = 60.

(c) Determine an estimate for F(63) by employing the local linearization.

(d) Do you think your estimate in (c) is too large or too small? Why?

3. An object moving along a straight line path has a differentiable position function y = s(t); s(t) measures the object’s position relative to the origin at time t. It is known that at time t = 9 seconds, the object’s position is s(9) = 4 feet (i.e., 4 feet to the right of the origin). Furthermore, the object’s instantaneous velocity at t = 9 is −1.2 feet per second, and its acceleration at the same instant is 0.08 feet per second per second.

(a) Use local linearity to estimate the position of the object at t = 9.34.

(b) Is your estimate likely too large or too small? Why?

(c) In everyday language, describe the behavior of the moving object at t = 9. Is it moving toward the origin or away from it? Is its velocity increasing or decreasing? 85

4. For a certain function f , its derivative is known to be f 0 (x) = (x − 1)e −x 2 . Note that you do not know a formula for y = f (x).

(a) At what x-value(s) is f 0 (x) = 0? Justify your answer algebraically, but include a graph of f 0 to support your conclusion.

(b) Reasoning graphically, for what intervals of x-values is f 00(x) > 0? What does this tell you about the behavior of the original function f? Explain.

(c) Assuming that f (2) = −3, estimate the value of f (1.88) by finding and using the tangent line approximation to f at x = 2. Is your estimate larger or smaller than the true value of f (1.88)? Justify your answer.

An Introduction to the Mathematics of Financial Derivatives

An Introduction to the Mathematics of Financial Derivatives is a popular, intuitive text that eases the transition between basic summaries of financial engineering to more advanced treatments using stochastic calculus. Requiring only a basic knowledge of calculus and probability, it takes readers on a tour of advanced financial engineering. This classic title has been revised by Ali Hirsa, who accentuates its well-known strengths while introducing new subjects, updating others, and bringing new continuity to the whole. Popular with readers because it emphasizes intuition and common sense, An Introduction to the Mathematics of Financial Derivatives remains the only "introductory" text that can appeal to people outside the mathematics and physics communities as it explains the hows and whys of practical finance problems.

An Introduction to the Mathematics of Financial Derivatives is a popular, intuitive text that eases the transition between basic summaries of financial engineering to more advanced treatments using stochastic calculus. Requiring only a basic knowledge of calculus and probability, it takes readers on a tour of advanced financial engineering. This classic title has been revised by Ali Hirsa, who accentuates its well-known strengths while introducing new subjects, updating others, and bringing new continuity to the whole. Popular with readers because it emphasizes intuition and common sense, An Introduction to the Mathematics of Financial Derivatives remains the only "introductory" text that can appeal to people outside the mathematics and physics communities as it explains the hows and whys of practical finance problems.

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ML Aggarwal Limits and Derivatives ISC Class-11 Maths Understanding Ch-13

Limits Definition

A limit of a function f(x) is defined as a value, where the function reaches as the limit reaches some value. Limits are used to define integration, integral calculus and continuity of the function.

If f(y) is a function, then the limit of the function can be represented as

Limits

l is called the limit of the function f(x) if the equation is given as x → a, f(x) → l, and this is symbolically written for all the limits, the function should assume at a given point x = a. x could approach a number in two ways, either from the left or from the right, i.e., all the values of x near a could be greater than a or could be less than a.

Right-Hand Value

In this type of limits, Right-hand limit Value is referred to the situation in which f(x) gets dictated by values of f(x) when x tends to from the right.

What is a Left-Hand Value?

In the case of Left-hand limit, when x tends to from the left, the value of f(x) gets dictated by values of f(x).

In this case, the right and left-hand limits are different, and hence we say that the limit of f(x) as x tends to zero does not exist (even though the function is defined at 0). This could also be followed in limits and continuity of values.

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ML Aggarwal Limits and Derivatives ISC Class-11 Maths Understanding Ch-13

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ML Aggarwal Limits and Derivatives ISC Class-11 Maths Understanding Ch-13

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ML Aggarwal Limits and Derivatives ISC Class-11 Maths Understanding Ch-13

Chapter Test

ML Aggarwal Limits and Derivatives ISC Class-11 Maths Understanding Ch-13

-: End of Limits and Derivatives ISC Class-11 ML Aggarwal Maths Understanding Chapter-13 Solution :-

Detailed Solutions to the Above Questions

Solution to Question 1
For function ( f ) to take real values, the expression under the radical in the numerator must be non negative, and the expression under the radical in the denominator must be positive hence the inequalities to solve
( x - 1 ge 0 ) and ( 4 - x^2 gt 0 )
The solution sets for the first and second inequalities are respectively
( x ge 1 ) and ( -2 lt x lt 2 )
Both inequalities must be satisfied simultaneously, therefore the domain of the given function is the intersection of the sets ( x ge 1 ) and ( -2 lt x lt 2 ) which is given by
( 1 le x lt 2 )

b)
The limit is of the indeterminate form ( dfrac ).
Divide all terms in the numerator and all terms of the denominator by with the term with the highest power which in ( x^4 )
( lim_ dfrac<-x^3+2x-1> = lim_ dfrac+dfrac<2x>-dfrac<1>> - dfrac<3 x^3> + dfrac<9> >).
Simplify rational terms
( = lim_ dfrac+dfrac<2>-dfrac<1>><1 - dfrac<3> + dfrac<9> > = dfrac<0+0-0> <1 - 0 + 0 >= dfrac<0> <1>= 0)

c)
The limit is of the indeterminate form ( infty cdot 0 ).
Let ( t = dfrac<3> ) and rewrite the limit in terms of t.
( lim_ x sin(dfrac<3>) = lim_ 3 dfrac ).
Using the well known result ( lim_ dfrac = 1 ), the limit evaluates to
( = 3 imes 1 = 3)

e)
( lim_ dfrac ).
It is well known that the range of ( sin(x) ) is given by
( -1 le sin(x) le 1 )
Add 1 to all terms of the inequality to obtain the following inequality
( -1 + 1 le sin(x) + 1 le 1 + 1 )
( 0 le sin(x) + 1 le 2 )
Divide all terms of the above inequality by positive ( x )
( dfrac<0> le dfrac le dfrac<2> )
We have ( lim_ dfrac<0> = 0 ) and ( lim_ dfrac<2> = 0 )
Using the squeezing (or Sandwich) theorem, we can evaluate the given limit as follows
( lim_ dfrac = 0 )  ( m = e^<1-1>= 1 )
The equation of the tangent line at the point ( (1,1) ) is given by
( y - 1 = 1 imes (x - 1) )
Which simplifies to
( y = x )
Hence the graphs of ( y = e^ ) and ( y = x ) are tangent at the point ( (1,1) ) and we can therefore state graphically that ( e^ ge x ).
b)
The first and secopnd derivatives of ( f ) are given by
( f'(x) = dfrac<2>-e^ )
( f''(x) = x-e^ )
A point of inflection occurs at a value of ( x ) where ( f''(x) ) change sign. We have seen in part a) that ( e^ ge x ) which may be written as
( x-e^ le 0 )
and therefore ( f''(x) ) is negative and has a zero at ( x = 1 ). Hence the graph of ( f(x) ) is concave down and does not have a point of inflection because ( f''(x) ) does not change sign.

d) Let ( u = sqrt + 2> ) , write function (m ) as (m = sin u ) then use the chain rule of derivatives

e)
Rewrite ( 3^ < 2x+3>) and ( log_3(2x-1) ) as
( 3^ <2x+3>= e^<(2x+3) ln 3>) , change of base of exponentials
( log_3(2x-1) = dfrac< ln(2x-1)>< ln 3>) , change of base of logarithms
Substitute and rewrite ( n(x) ) as
( n(x) = 3^ < 2x+3>+ log_3(2x-1) = e^ <(2x+3) ln 3>+ dfrac< ln(2x-1)> < ln 3>)
We now calculate the derivative
( n'(x) = ( 2 ln 3 ) e^ <(2x+3) ln 3>+ dfrac<1> < ln 3>dfrac<2> <2x-1>= ( 2 ln 3 ) 3^ <2x+3>+ dfrac<2> )

ML Aggrawal ISC Class-11 APC Understanding Mathematics Solutions

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Differentiation allows us to find rates of change. For example, it allows us to find the rate of change of velocity with respect to time (which is acceleration). It also allows us to find the rate of change of x with respect to y, which on a graph of y against x is the gradient of the curve. There are a number of simple rules which can be used to allow us to differentiate many functions easily.

If y = some function of x (in other words if y is equal to an expression containing numbers and x's), then the derivative of y (with respect to x) is written dy/dx, pronounced "dee y by dee x" .

Differentiating x to the power of something

2) If y = kx n , dy/dx = nkx n-1 (where k is a constant- in other words a number)

Therefore to differentiate x to the power of something you bring the power down to in front of the x, and then reduce the power by one.

If y = x 4 , dy/dx = 4x 3
If y = 2x 4 , dy/dx = 8x 3
If y = x 5 + 2x -3 , dy/dx = 5x 4 - 6x -4

This looks hard, but it isn't. The trick is to simplify the expression first: do the division (divide each term on the numerator by 3x ½ . We get:
(1/3)x 3/2 + (5/3)x ½ - x -½ (using the laws of indices).

So differentiating term by term: ½ x ½ + (5/6)x -½ + ½x -3/2 .

There are a number of ways of writing the derivative. They are all essentially the same:

(1) If y = x 2 , dy/dx = 2x
This means that if y = x 2 , the derivative of y, with respect to x is 2x.

(2) d (x 2 ) = 2x
dx
This says that the derivative of x 2 with respect to x is 2x.

(3) If f(x) = x 2 , f'(x) = 2x
This says that is f(x) = x 2 , the derivative of f(x) is 2x.

Finding the Gradient of a Curve

A formula for the gradient of a curve can be found by differentiating the equation of the curve.

What is the gradient of the curve y = 2x 3 at the point (3,54)?
dy/dx = 6x 2
When x = 3, dy/dx = 6× 9 = 54

Teaching Calculus

A very typical calculus problem is given the equation of a function, to find information about it (extreme values, concavity, increasing, decreasing, etc., etc.). This is usually done by computing and analyzing the first derivative and the second derivative. All the textbooks show how to do this with copious examples and exercises. I have nothing to add to that. One of the “tools” of this approach is to draw a number line and mark the information about the function and the derivative on it.

A very typical AP Calculus exam problem is given the graph of the derivative of a function, but not the equation of either the derivative or the function, to find all the same information about the function. For some reason, student find this difficult even though the two-dimensional graph of the derivative gives all the same information as the number line graph and, in fact, a lot more.

Looking at the graph of the derivative in the x,y-plane it is easy to very determine the important information. Here is a summary relating the features of the graph of the derivative with the graph of the function.

 Feature the function > 0 is increasing < 0 is decreasing changes – to + has a local minimum changes + to – has a local maximum increasing is concave up decreasing is concave down extreme value has a point of inflection

Here’s a typical graph of a derivative with the first derivative features marked. Here is the same graph with the second derivative features marked. The AP Calculus Exams also ask students to “Justify Your Answer.” The table above, with the columns switched does that. The justifications must be related to the given derivative, so a typical justification might read, “The function has a relative maximum at x = -2 because its derivative changes from positive to negative at x = -2.”

Why is e ^( pi i ) = -1?

I was watching an episode of The Simpsons the other day, the one where Homer gets sucked into the third dimension, and in this 3-D world, there was an equation that said . So I put it into the calculator and it worked, but I have no idea why, because e to any power isnt supposed to be a negative number, and I thought pi was in no way related to e .

If you could explain the process, it would save lots of time pondering and plugging e , pi , and i into the calculator in random ways to figure out whats going on.

The first question to ask, though, is not "why does ", but rather, "what does even mean ?" In other words, what does it mean to raise a number to an imaginary power?

Once that question is answered, it will be much more clear why . It turns out that for all x , a fact which is known as de Moivre's formula , and illustrates how closely related the exponential function is to the trigonometric functions. From this formula, it follows immediately that .

So now, the question is, why is the "right" thing to define what e raised to an imaginary power means?

Raising a number to an imaginary power makes no sense based on the original definition of exponentiation you learned, where means " a multiplied by itself b times." That definition only makes sense when b is a positive integer. After all, what would it mean to multiply something by itself i times??

Of course, the original definition doesn't even make sense for fractions and negative numbers. You should have learned how to extend the definition to include fractions. For example, since 1/3 is that number which, when multiplied by 3, gives you 1, it makes sense to define to be that number which, if you raise it to the power of 3, would give you (i.e., a ) in other words, is defined to be the cube root of a . Similarly, you learned how to extend the definition to negative exponents by .

But none of these considerations give any clue as to what raising a number to a complex power should mean. Instead, we need to express exponentiation, or its properties, in some way that can be extended to complex powers.

The first way to do this is to use the fact that happens to be equal to the infinite sum

(where n ! means n factorial, the product of the numbers 1,2,. . . , n ).

The reason why this is so depends on the theory of Taylor series from calculus, which would take too long to describe here. You will encounter it in a calculus class at some point, if you haven't already.

Now, this infinite sum makes perfectly good sense even for imaginary numbers. By plugging in ix in place of x , you get

Now it turns out that is the infinite sum for cos x , while is the infinite sum for sin x (again by the theory of Taylor series). Therefore, .

Now, this may be a little unsatisfying to you since I haven't explained why , cos x , and sin x equal those three different infinite sums. I can't do so without assuming some calculus background that you may not have.

However, here's another way of understanding why . It too involves some calculus, but I can describe the calculus involved more easily.

Associated to many functions f ( x ) is another function f '( x ), called the derivative of f ( x ). It measures how rapidly f ( x ) is changing at the value x .

If , f ( x ) may represent, for example, an exponentially growing population. The rate of change of such a population (the number of births per day, for example) is directly proportional to the current size of the population that is, f '( x ) is a constant times f ( x ). When , that constant is exactly 1 (that's the property which defines the number e ). More generally, if , then .

What about the trigonometric functions? Well, if f ( x ) = sin x , then f '( x ) = cos x , and if f ( x ) = cos x , then f '( x ) = - sin x .

If you think about it for a minute, these equations are very reasonable. First of all, when x =0, sin x equals zero but increases as x increases in fact, the slope of the graph of y = sin x at the point (0,0) is 1, which is another way of saying that the rate of increase there is 1, so f '(0) = 1.

But then, as x increases to , the rate of increase drops off and eventually sin x stops increasing altogether and starts decreasing. In other words, f '( x ) drops to zero when , and becomes -1 by the time x reaches pi (see the picture).

Therefore, f '( x ) is a function which starts at 1 when x =0, decreases to 0 when , drops to -1 when , rises back to 0 when , and so on. This is precisely what the cosine function does, so it should be no surprise that f '( x ) = cos x . Similar reasoning shows why it is reasonable that, when f ( x )=cos x , f '( x )=-sin x . The exact proofs of these facts you will see in a calculus class.

Now, keeping those facts in mind, what should be? If we write it in terms of real and imaginary parts g ( x ) + i h ( x ), what should the functions g ( x ) and h ( x ) be?

The key is to take the derivative. It is only reasonable to define in such a way that it still has the same properties as mentioned above, namely, the derivative of should still equal . Therefore, if , we should have

But f '( x ) should also equal g '( x ) + i h '( x ), so we are looking for a pair of functions g and h for which h ' = g and g ' = - h . This is exactly the same interrelationship that the sine and cosine functions have, as we saw above. It also turns out that these two equations, together with the conditions g (0)=1 and h (0)=0 that arise from the fact that needs to equal 1, uniquely determine the functions g and h .

It follows from all this that g must be the cosine function and h must be the sine function. That is why .

Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences

The derivative of a function of a single variable tells us how quickly the value of the function changes as the value of the independent variable changes. Intuitively, it tells us how “steep” the graph of the function is. We might wonder if there is a similar idea for graphs of functions of two variables, that is, surfaces. It is not clear that this has a simple answer, nor how we might proceed. We will start with what seem to be very small steps toward the goal. Surprisingly, it turns out that these simple ideas hold the keys to a more general understanding.

Subsection 7.3.1 First-Order Partial Derivatives

The derivative of a single-variable function (f(x)) tells us how much (f(x)) changes as (x) changes. There is no ambiguity when we speak about the rate of change of (f(x)) with respect to (x) since (x) must be constrained to move along the (x)-axis. The situation becomes more complicated, however, when we study the rate of change of a function of two or more variables. The obvious analogue for a function of two variables (g(x,y)) would be something that tells us how quickly (g(x,y)) increases as (x) and (y) increase. However, in most cases this will depend on how quickly (x) and (y) are changing relative to each other.

Example 7.15 . Analyzing a Simple Rate of Change in One Variable.

Given (f(x,y)=y^2 ext<,>) analyze the rate of change of (f) when one variable changes and the other variable is kept constant.

If we look at a point ((x,y,y^2)) on this surface, the value of a function does not change at all if we fix (y) and let (x) increase, but increases like (y^2) if we fix (x) and let (y) increase.

Example 7.16 . Analyzing Rates of Change in One Variable.

Given (f(x,y)=x^2+y^2 ext<,>) analyze the rate of change of (f) when one variable changes and the other variable is kept constant.

Now let us consider what happens to (f(x,y)) when both (x) and (y) are increasing, perhaps at different rates. We can think of this as being a movement in a certain direction of a point in the (x)-(y)-plane. A point and a direction defines a line in the (x)-(y)-plane, and so we are asking how the function changes as we move along this line.

Introducing cross-sections: Let us then imagine a plane perpendicular to the (x)-(y)-plane that intersects the (x)-(y)-plane along this line. This plane will intersect the surface of (f) in a curve, so we can just look at the behaviour of this curve in the given plane.

Figure 7.3 shows the plane (x+y=1 ext<,>) which is the plane perpendicular to the line (x+y=1) in the (x)-(y)-plane. Observe that its intersection with the surface of (f) is a curve, in fact, a parabola. We will refer to such a curve as the cross-section of the surface above the line in the (x)-(y)-plane.

How to think of a rate of change: We can now look at the rate of change (or slope) of (f) in a particular direction by looking at the slope of a curve in a plane — something we already have experience with.

Simple case - lines parallel to the x- or y- axis: Let's start by looking at some particularly easy lines: Those parallel to the (x)- or (y)- axis. Suppose we are interested in the cross-section of (f(x,y)) above the line (y=b ext<.>) If we substitute (b) for (y) in (f(x,y) ext<,>) we get a function in one variable, describing the height of the cross-section as a function of (x ext<.>) Because (y=b) is parallel to the (x)-axis, if we view it from a vantage point on the negative (y)-axis, we will see what appears to be simply an ordinary curve in the (x)-(z)-plane.

We now consider a particular cross-section. The cross-section above the line (y=2) consists of all points ((x,2,x^2+4) ext<.>) Looking at this cross-section we see what appears to be just the curve (f(x)=x^2+4 ext<.>) At any point on the cross-section, ((a,2,a^2+4) ext<,>) the slope of the surface in the direction of the line (y=2) is simply the slope of the curve (f(x)=x^2+4 ext<,>) namely (2x ext<.>) Figure 7.4 shows the same parabolic surface as before, but now cut by the plane (y=2 ext<.>) The left graph shows the cut-off surface, the right shows just the cross-section.

If, for example, we're interested in the point ((-1,2,5)) on the surface, then the slope in the direction of the line (y=2) is (2x=2(-1)=-2 ext<.>) This means that starting at ((-1,2,5)) and moving on the surface, above the line (y=2 ext<,>) in the direction of increasing (x)-values, the surface goes down of course moving in the opposite direction, toward decreasing (x)-values, the surface will rise.

Generalizing our findings: If we're interested in some other line (y=k ext<,>) there is really no change in the computation. The equation of the cross-section above (y=k) is (x^2+k^2) with derivative (2x ext<.>) We can save ourselves the effort, small as it is, of substituting (k) for (y ext<:>) all we are in effect doing is temporarily assuming that (y) is some constant. With this assumption, the derivative ((x^2+y^2)=2x ext<.>) To emphasize that we are only temporarily assuming (y) is constant, we use a slightly different notation: ((x^2+y^2)=2x ext<>) the “(partial)” reminds us that there are more variables than (x ext<,>) but that only (x) is being treated as a variable. We read the equation as “the partial derivative of ((x^2+y^2)) with respect to (x) is (2x ext<.>)”

Based on this discussion, we are now in a position to formally introduce the . Of course, we can do the same sort of calculation for lines parallel to the (y)-axis. We temporarily hold (x) constant, which gives us the equation of the cross-section above a line (x=k ext<.>) We can then compute the derivative with respect to (y ext<>) this will measure the slope of the curve in the (y) direction.

Definition 7.17 . First-Order Partial Derivatives of (f(x,y)).

Suppose (f(x,y)) is a two-variable function.

Then, the at ((x,y)) is given by

provided the limit exists,

and the at ((x,y)) is given by

provided the limit exists.

Note: Convenient alternate notations for the partial derivatives of (z=f(x,y)) are given by

Example 7.18 . Partial Derivative with respect to (y).

Find the partial derivative with respect to (y) of (f(x,y)=sin(xy)+3xy ext<.>)

The partial derivative with respect to (y) of (f(x,y)=sin(xy)+3xy) is

Example 7.19 . Partial Derivatives.

Given the two-variable function (f(x,y)=x^2-xy+y^5 ext<,>)

Find the first-order partial derivatives (f_x) and (f_y ext<.>)

Calculate (f_x(2,1)) and interpret your result.

Calculate (f_y(2,1)) and interpret your result.

To find the first-order partial derivative of (f) with respect to (x ext<,>) we treat the variable (y) as a constant and then differentiate as usual. That is,

Next, to compute the first-order partial derivative of (f) with respect to (y ext<,>) we treat the variable (x) as a constant. Then,

Letting (x=2) and (y=1 ext<,>) we find

This means that (f) is increasing at a rate of 3 in the (x)-direction at the point ((2,1) ext<.>)

We again let (x=2) and (y=1 ext<.>) Then,

which gives the rate at which (f) is changing in the (y)-direction at the point ((2,1) ext<.>)

Example 7.20 . First-Order Partial Derivatives.

Find the first-order partial derivatives of the following functions.

(g(s,t) = left(s + st^2 -t ^3 ight)^2 ext<.>)

We first compute (f_x ext<:>) to do this, we need to treat (y) as a constant. Then

Now to compute (f_y ext<,>) we consider (x) to be constant, which gives

Treating the variable (t) as a constant and differentiating with respect to (s) gives

And so if we treat (s) as a constant and differentiate in the (t)-direction, we obtain

Differentiating in the (x)-direction gives

and differentiating in the (y)-direction gives

Example 7.21 . First-Order Partial Derivatives.

Find the first-order partial derivatives of the function

Since (f) is a three-variable function, it will have three first-order partial derivatives:

So to compute the derivative in the (x)-direction, we need to consider both (y) and (z) as constants and then differentiate with respect to (x ext<,>) and similarly for the derivatives in the (y)- and (z)-direction. We then obtain

Subsection 7.3.2 Applications

In economics, the is a specific form of that yields the amount of output produced by the amounts of two or more inputs, which are in technological relationship among each other. Here, we only consider one type of example, namely the connection between physical capital and labour, which are defined as follows.

Definition 7.22 . Cobb-Douglas Production Function.

The Cobb-Douglas production function is defined by

to represent the amount of output, (Y ext<,>) produced by two inputs that are in technological relationship to each other, where specifically

(L) is the total number of person-hours worked per year

(K) is the real value of all physical capital such as machinery, equipment, and buildings

(a) is the total factor productivity and

(bin (0,1)) and (1-b) are the output elasticities of labour and capital respectively.

The partial derivative (Y_L) measures the rate of change of production with respect to the amount of money expended for labour, when the level of capital expenditure is held fixed. Therefore, (Y_L) is called the . Similarly, the partial derivative (Y_K) measures the rate of change of production with respect to the amount of money expended for capital, when the level of labour expenditure is held constant. Therefore, (Y_K) is called the .

We showcase these concepts with the following example.

Example 7.23 . Marginal Productivity.

Suppose a country's production can be described by the function

units, where (L) is units of labour and (K) is units of capital.

Calculate the marginal productivity of labour and the marginal productivity of capital.

Evaluate (Y_L) and (Y_K) when (L=100) and (K=55 ext<.>)

The marginal productivity of labour is calculated to be

and the marginal productivity of capital is

When (L=100) and (K=55 ext<,>) we have

We see that the marginal productivity of labour is less than the marginal productivity of capital when 100 units is spent on labour and 55 units is spent on capital. Therefore, if we increase the amount spent on capital while keeping the amount spent on labour constant, the resulting productivity will be higher than if we were to increase the amount spent on labour instead.

A further application in economics of partial derivatives is to two-variable functions that represent the of two commodities that are either competitive or complementary.

The relative demands of two commodities are defined as follows:

Definition 7.24 . Relative Demand Equations of Two Commodities.

Let (A) and (B) be two commodities, then the are given by

Definition 7.25 . Competitive Commodities.

The relative demands of two commodities are termed , if they have an inverse relationship to each other, i.e. an increase in price (or decrease in demand) of one commodity results in an increase in demand of the other commodity.

Examples of competitive commodities are bread and cereal, beer and wine, or cherries and strawberries.

Definition 7.26 . Complementary Commodities.

The relative demands of two commodities are termed , if they have a direct relationship to each other, i.e. an increase in demand of one commodity results in an increase in demand of the other commodity and vice versa.

Examples of complementary commodities are flat screen TVs and blue-ray players, hamburgers and hamburger buns, or apple pie and vanilla ice cream.

Using partial derivatives, we can readily test whether two relative demands are competitive or complementary.

Definition 7.27 . Partial Derivatives of Competitive and Complementary Commodities.

then the commodities (A) and (B) are competitive.

then the commodities (A) and (B) are complementary.

Example 7.28 . Competitive and Complementary Commodities.

Let (q_A) denote the quantity of apples demanded and (q_B) denote the quantity of blueberries demanded in a supermarket. If the demand equations for apples and blueberries are given by

respectively, determine whether the products are competitive, complementary or neither.

Since (p_A) and (p_B) must be positive, we determine that

We thus conclude that the two products are competitive.

Example 7.29 . Competitive and Complementary Commodities.

A retailer sells both raincoats and umbrellas. Let (q_A) denote the quantity of umbrellas demanded and (q_B) denote the quantity of raincoats demanded. The demand equations for raincoats and umbrellas are, respectively,

Determine whether these two commodities are competitive, complementary, or neither.

As always, (p_A) and (p_B) must be positive. We thus determine that

and that the two products are competitive.

Subsection 7.3.3 Tangent Plane

So far, using no new techniques, we have succeeded in measuring the slope of a surface in two quite special directions. For functions of one variable, the derivative is closely linked to the notion of tangent line. For surfaces, the analogous idea is the tangent plane—a plane that just touches a surface at a point, and has the same slope as the surface in all directions. Even though we haven't yet figured out how to compute the slope in all directions, we have enough information to find tangent planes. Suppose we want the plane tangent to a surface at a particular point ((a,b,c) ext<.>) If we compute the two partial derivatives of the function for that point, we get enough information to determine two lines tangent to the surface, both through ((a,b,c)) and both tangent to the surface in their respective directions. These two lines determine a plane, that is, there is exactly one plane containing the two lines: the tangent plane. Figure 7.5 shows (part of) two tangent lines at a point, and the tangent plane containing them.

How can we discover an equation for this tangent plane? We know a point on the plane, ((a,b,c) ext<>) we need a vector normal to the plane. If we can find two vectors, one parallel to each of the tangent lines we know how to find, then the cross product of these vectors will give the desired normal vector.

How can we find vectors parallel to the tangent lines? Consider first the line tangent to the surface above the line (y=b ext<.>) A vector (langle u,v,w angle) parallel to this tangent line must have (y) component (v=0 ext<,>) and we may as well take the (x) component to be (u=1 ext<.>) The ratio of the (z) component to the (x) component is the slope of the tangent line, precisely what we know how to compute. The slope of the tangent line is (f_x(a,b) ext<,>) so

In other words, a vector parallel to this tangent line is (langle 1,0,f_x(a,b) angle ext<,>) as shown in Figure 7.6. If we repeat the reasoning for the tangent line above (x=a ext<,>) we get the vector (langle 0,1,f_y(a,b) angle ext<.>)

Now to find the desired normal vector we compute the cross product, (langle 0,1,f_y angle imeslangle 1,0,f_x angle= langle f_x,f_y,-1 angle ext<.>) From our earlier discussion of planes, we can write down the equation we seek: (f_x(a,b)x+f_y(a,b)y-z=k ext<,>) and (k) as usual can be computed by substituting a known point: (f_x(a,b)(a)+f_y(a,b)(b)-c=k ext<.>) There are various more-or-less nice ways to write the result:

Example 7.30 . Tangent Plane to a Sphere.

Find the plane tangent to (x^2+y^2+z^2=4) at ((1,1,sqrt2) ext<.>)

The point ((1,1,sqrt2)) is on the upper hemisphere, so we use (ds f(x,y)=sqrt<4-x^2-y^2> ext<.>) Then (ds f_x(x,y)=-x(4-x^2-y^2)^<-1/2>) and (ds f_y(x,y)=-y(4-x^2-y^2)^<-1/2> ext<,>) so (f_x(1,1)=f_y(1,1)=-1/sqrt2) and the equation of the plane is

The hemisphere and this tangent plane are pictured in Figure 7.5.

So it appears that to find a tangent plane, we need only find two quite simple ordinary derivatives, namely (f_x) and (f_y ext<.>) This is true if the tangent plane exists. It is, unfortunately, not always the case that if (f_x) and (f_y) exist there is a tangent plane. Consider the function (xy^2/(x^2+y^4)) with (f(0,0)) defined to be 0, pictured in Figure 7.1. This function has value 0 when (x=0) or (y=0 ext<.>) Now it's clear that (f_x(0,0)=f_y(0,0)=0 ext<,>) because in the (x) and (y) directions the surface is simply a horizontal line. But it's also clear from the picture that this surface does not have anything that deserves to be called a tangent plane at the origin, certainly not the (x)-(y)-plane containing these two tangent lines.

When does a surface have a tangent plane at a particular point? What we really want from a tangent plane, as from a tangent line, is that the plane be a “good” approximation of the surface near the point. Here is how we can make this precise:

Definition 7.31 . Tangent Plane.

Let (Delta x=x-x_0 ext<,>) (Delta y=y-y_0 ext<,>) and (Delta z=z-z_0) where (z_0=f(x_0,y_0) ext<.>) The function (z=f(x,y)) is differentiable at ((x_0,y_0)) if

where both (epsilon_1) and (epsilon_2) approach 0 as ((x,y)) approaches ((x_0,y_0) ext<.>)

This definition takes a bit of absorbing. Let's rewrite the central equation a bit:

The first three terms on the right are the equation of the tangent plane, that is,

is the (z)-value of the point on the plane above ((x,y) ext<.>) Equation (7.3.1) says that the (z)-value of a point on the surface is equal to the (z)-value of a point on the plane plus a “little bit,” namely (epsilon_1Delta x + epsilon_2Delta y ext<.>) As ((x,y)) approaches ((x_0,y_0) ext<,>) both (Delta x) and (Delta y) approach 0, so this little bit (epsilon_1Delta x + epsilon_2Delta y) also approaches 0, and the (z)-values on the surface and the plane get close to each other. But that by itself is not very interesting: since the surface and the plane both contain the point ((x_0,y_0,z_0) ext<,>) the (z) values will approach (z_0) and hence get close to each other whether the tangent plane is tangent to the surface or not. The extra condition in the definition says that as ((x,y)) approaches ((x_0,y_0) ext<,>) the (epsilon) values approach 0—this means that (epsilon_1Delta x + epsilon_2Delta y) approaches 0 much, much faster, because (epsilon_1Delta x) is much smaller than either (epsilon_1) or (Delta x ext<.>) It is this extra condition that makes the plane a tangent plane.

We can see that the extra condition on (epsilon_1) and (epsilon_2) is just what is needed if we look at partial derivatives. Suppose we temporarily fix (y=y_0 ext<,>) so (Delta y=0 ext<.>) Then the equation from the definition becomes

Now taking the limit of the two sides as (Delta x) approaches 0, the left side turns into the partial derivative of (z) with respect to (x) at ((x_0,y_0) ext<,>) or in other words (f_x(x_0,y_0) ext<,>) and the right side does the same, because as ((x,y)) approaches ((x_0,y_0) ext<,>) (epsilon_1) approaches 0. Essentially the same calculation works for (f_y ext<.>)

Subsection 7.3.4 Second-Order Partial Derivatives

The first-order partial derivatives (frac) and (frac) of a two-variable function (f(x,y)) are again functions of the two variables (x) and (y ext<.>) Therefore, we can compute the derivative of each function (frac(x,y)) and (frac(x,y)) to obtain the of (f ext<.>)

Definition 7.32 . Second-Order Partial Derivatives of (f(x,y)).

Suppose (f(x,y)) is a two-variable function. Then we obtain the four by differentiating the functions (f_x) and (f_y) with respect to (x) and (y ext<:>)

The notation for the second-order partial derivative is analogous to the notation for the ordinary second derivative in single-variable calculus:

When multiple distinct independent variables are involved in second-order partial derivatives, then they are also referred to as . For example, (f_) and (f_ ext<.>)

In general, we have that (f_ eq f_ ext<.>) However, if the second-order partial derivatives are continuous around a point ((a,b) ext<,>) then (f_(a,b) = f_(a,b) ext<.>) This is a theorem that is variously referenced to as Schwarz's theorem or Clairaut's theorem, but we would need real analysis in order to prove it, so we state it here without proof. Most functions we will encounter in this course satisfy Clairaut's theorem.

Figure 7.7 is a schematic way of demonstrating how the four second-order partial derivatives of (f) are calculated.

Theorem 7.33 . Clairaut's Theorem.

If the mixed partial derivatives are continuous, they are equal.

Example 7.34 . Second-Order Partial Derivatives.

Find the second-order partial derivatives of the function

We will first calculate (f_x) and (f_y ext<.>)

Hence, the four second-order partial derivatives of (f) are

The first-order partial derivatives of (g) are

Thus, the second-order partial derivatives of (g) are given by

Exercises for Section 7.3.
Exercise 7.3.1 .

Find (f_x) and (f_y) of the following functions (f(x,y) ext<.>)

Given (f(x,y) = cos(x^2y) + y^3 ext<,>) we compute the following partial derivatives:

Given (f(x,y) = dfrac ext<,>) we compute the following partial derivatives:

Given (f(x,y)=e^ ext<,>) we compute the following partial derivatives:

Given (f(x,y) = xyln(xy) ext<,>) we compute the following partial derivatives:

Given (f(x,y) = sqrt<1-x^2-y^2> ext<,>) we compute the following partial derivatives:

Given (f(x,y) = x an(y) ext<,>) we compute the following partial derivatives:

Given (f(x,y) = dfrac<1> ext<,>) we compute the following partial derivatives:

Exercise 7.3.2 .

Evaluate the first partial derivatives of the function at the given point.

Let us now apply the knowledge of these rules of calculus in our original equation and find the derivative of the Cost Function w.r.t to both ‘m’ and ‘b’. Revising the Cost Function equation : For simplicity, let us get rid of the summation sign. The summation part is important, especially with the concept of Stochastic gradient descent (SGD ) vs. batch gradient descent. During the batch gradient descent, we look at the error of all the training examples at once, while in the SGD, we look at each error at a time. However, to keep things simple, we will assume that we are looking at each error one at a time. Now let’s calculate the gradient of Error w.r.t to both m and b : Plugging the values back in the cost function and multiplying it with the learning rate:  This 2 in this equation isn’t that significant since it just says that we have a learning rate twice as big or half as big. So let’s get rid of it too. So, Ultimately, this entire article boils down to two simple equations representing the equations for Gradient Descent. m¹,b¹ = next position parameters m⁰,b⁰ = current position parameters

Hence,to solve for the gradient, we iterate through our data points using our new m and b values and compute the partial derivatives. This new gradient tells us the slope of our cost function at our current position and the direction we should move to update our parameters. The size of our update is controlled by the learning rate.